Putnam All Questions

7/26/2019 Putnam All Questions

1/109

The Forty-Sixth Annual William Lowell Putnam Competition

Saturday, December 7, 1985

A1 Determine, with proof, the number of ordered triples

of sets which have the property that

(i)

,and

(ii)

.

Express your answer in the form

, where

are nonnegative integers.

A2 Let be an acute triangle. Inscribe a rectangle in

with one side along a side of . Then inscribe a rectan-

gle in the triangle formed by the side of

opposite

the side on the boundary of , and the other two sides of

, with one side along the side of

. For any polygon

, let

denote the area of

. Find the maximumvalue, or show that no maximum exists, of

,

where ranges over all triangles and

over all rect-

angles as above.

A3 Let

be a real number. For each integer

, define

a sequence

,

by the condition

Evaluate

.

A4 Define a sequence

by

and

for

. Which integers between 00 and 99 inclusive oc-

cur as the last two digits in the decimal expansion of

infinitely many ?

A5 Let

. For

which integers ,

is

?

A6 If

is a polynomial with

real coefficients , then set

Let

. Find, with proof, a polyno-

mial

with real coefficients such that

(i)

, and

(ii)

for every integer

.

B1 Let be the smallest positive integer for which there ex-ist distinct integers

such that the

polynomial

has exactly nonzero coefficients. Find, with proof, a

set of integers

for which this min-

imum is achieved.

B2 Define polynomials

for

by

,

for

, and

for

. Find, with proof, the explicit factorization of

into powers of distinct primes.

B3 Let

......

.... . .

be a doubly infinite array of positive integers, and sup-

pose each positive integer appears exactly eight times

in the array. Prove that

for some pair ofpositive integers

.

B4 Let be the unit circle

. A point is chosen

randomly on the circumference and another point

is chosen randomly from the interior of (these points

are chosen independently and uniformly over their do-

mains). Let be the rectangle with sides parallel to the

and

-axes with diagonal . What is the probability

that no point of lies outside of ?

B5 Evaluate

. You may assume

that

.

B6 Let be a finite set of real

matrices

,

, which form a group under matrix multi-

plication. Suppose that

, where

denotes the trace of the matrix

. Prove that

is the

zero matrix.


2/109

The Forty-Seventh Annual William Lowell Putnam Competition


A1 Find, with explanation, the maximum value off(x) =x3 3xon the set of all real numbersxsatisfyingx4 +36

13x2.

A2 What is the units (i.e., rightmost) digit of

1020000

10100 + 3

?

A3 Evaluate

n=0Arccot(n2+n+1), where Arccot tfor

t 0denotes the number in the interval0< /2withcot = t.

A4 A transversalof annnmatrixAconsists ofnentriesofA, no two in the same row or column. Let f(n) bethe number ofnn matricesAsatisfying the followingtwo conditions:

(a) Each entryi,j ofAis in the set {1, 0, 1}.(b) The sum of the n entries of a transversal is the

same for all transversals ofA.

An example of such a matrixA is

A=

1 0 10 1 0

0 1 0

.

Determine with proof a formula forf(n)of the form

f(n) =a1bn1 +a2bn2 +a3bn3 +a4,

where theais andbis are rational numbers.

A5 Suppose f1(x), f2(x), . . . , f n(x)are functions ofnrealvariablesx = (x1, . . . , xn) with continuous second-order partial derivatives everywhere on Rn. Suppose

further that there are constantscij such that

fixj

fjxi

=cij

for alli and j ,1 i n,1 j n. Prove that thereis a function g(x)on Rn such thatfi + g/xiis linearfor all i, 1 i n. (A linear function is one of theform

a0+a1x1+a2x2+ +anxn.)

A6 Let a1, a2, . . . , an be real numbers, and letb1, b2, . . . , bn be distinct positive integers. Sup-pose that there is a polynomial f(x) satisfying theidentity

(1 x)nf(x) = 1 +n

i=1

aixbi .

Find a simple expression (not involving any sums) fof(1) in terms ofb1, b2, . . . , bn andn (but independenofa1, a2, . . . , an).

B1 Inscribe a rectangle of base b and height h in a circlof radius one, and inscribe an isosceles triangle in th

region of the circle cut off by one base of the rectangl

(with that side as the base of the triangle). For whavalue ofh do the rectangle and triangle have the samarea?

B2 Prove that there are only a finite number of possibilitie

for the ordered tripleT = (x y, y z, z x), wherx,y,z are complex numbers satisfying the simultaneous equations

x(x 1) + 2yz = y(y 1) + 2zx = z(z 1) + 2xy,and list all such triplesT.

B3 Let consist of all polynomials inx with integer coefficients. Forf andg in and m a positive integer, lef g (mod m) mean that every coefficient offis an integral multiple ofm. Let n and p be positivintegers withp prime. Given thatf , g , h , rand s are i withrf+sg 1 (mod p) and f g h (mod p)prove that there exist F and G in with F f(modp),G g (mod p), andF G h (mod pn).

B4 For a positive real numberr, letG(r)be the minimumvalue of|r m2 + 2n2| for all integers m and nProve or disprove the assertion that limrG(r) exists and equals 0.

B5 Let f(x,y,z) = x2 + y2 + z2 + xyz . Lep(x,y,z), q(x,y,z), r(x,y,z) be polynomials witreal coefficients satisfying

f(p(x,y,z), q(x,y,z), r(x,y,z)) =f(x,y,z).

Prove or disprove the assertion that the sequencep,q,consists of some permutation ofx,y,z, where thnumber of minus signs is 0 or 2.

B6 SupposeA,B, C,D are n nmatrices with entries ia fieldF, satisfying the conditions thatABT andCDT

are symmetric and ADT BCT = I. HereI is thnnidentity matrix, and ifMis annnmatrix,MTis its transpose. Prove thatATD CTB= I.


3/109

The Forty-Eighth Annual William Lowell Putnam Competition


A1 Curves

and are defined in the plane as fol-

lows:

Prove that

.

A2 The sequence of digits

is obtained by writing the positive integers in order. If

the

-th digit in this sequence occurs in the part ofthe sequence in which the -digit numbers are placed,

define

to be . For example,

because

the 100th digit enters the sequence in the placement of

the two-digit integer 55. Find, with proof,

.

A3 For all real

, the real-valued function

satis-

fies

(a) If

for all real

, must

for all

real

? Explain.

(b) If

for all real

, must

for all

real

? Explain.A4 Let

be a polynomial, with real coefficients, in three

variables and be a function of two variables such that

for all real

and such that

,

, and

. Also let

be complex numbers

with

and

. Find

.

A5 Let

Prove or disprove that there is a vector-valued function

with the following properties:

(i)

have continuous partial derivatives for

all

;

(ii)

for all

;

(iii)

.

A6 For each positive integer

, let

be the number of

zeroes in the base 3 representation of

. For which pos-

itive real numbers

does the series

converge?

B1 Evaluate

B2 Let

and be integers with

,

and

. Prove that

B3 Let be a field in which

. Show that the set

of solutions to the equation

with

and

in is given by

and

where runs through the elements of such that

.

B4 Let

and let

and

for

. For each of

and

,

prove that the limit exists and find it or prove that thelimit does not exist.

B5 Let be the

-dimensional vector

. Let

be a

matrix of complex numbers such that

whenever

, with complex

,

not all zero, then at least one of the

is not real. Prove

that for arbitrary real numbers

, there are

complex numbers

such that

...

...

(Note: if

is a matrix of complex numbers,

isthe matrix whose entries are the real parts of the entries

of

.)

B6 Let be the field of

elements, where is an odd

prime. Suppose is a set of

distinct nonzero

elements of with the property that for each

in

, exactly one of and

is in . Let

be the num-

ber of elements in the intersection

.

Prove that

is even.


4/109

The Forty-Ninth Annual William Lowell Putnam Competition


A1 Let

be the region consisting of the points of thecartesian plane satisfying both

and

.

Sketch the region

and find its area.

A2 A not uncommon calculus mistake is to believe that the

product rule for derivatives says that . If

, determine, with proof, whether there exists

an open interval

and a nonzero function defined

on

such that this wrong product rule is true for

in

.

A3 Determine, with proof, the set of real numbers for

which

converges.

A4 (a) If every point of the plane is painted one of three

colors, do there necessarily exist two points of the

same color exactly one inch apart?

(b) What if three is replaced by nine?

A5 Prove that there exists auniquefunction from the set

of positive real numbers to

such that

and

for all

.

A6 If a linear transformation on an -dimensional vector

space has

eigenvectors such that any of them

are linearly independent, does it follow that is a scalar

multiple of the identity? Prove your answer.

B1 A composite (positive integer) is a product with

and not necessarily distinct integers in

.

Show that every composite is expressible as

, with

positive integers.

B2 Prove or disprove: If and are real numbers with

and

, then

.

B3 For every in the set

of positive inte-

gers, let

be the minimum value of

for all

nonnegative integers and

with

. Find, with

proof, the smallest positive real number with

for all

.

B4 Prove that if

is a convergent series of positive

real numbers, then so is

.

B5 For positive integers , let be the by

skew-symmetric matrix for which each entry in the first subdiagonals below the main diagonal is 1 and each

of the remaining entries below the main diagonal is -1.

Find, with proof, the rank of

. (According to one

definition, the rank of a matrix is the largest such that

there is a

submatrix with nonzero determinant.)

One may note that

B6 Prove that there exist an infinite number of ordered pairs

of integers such that for every positive integer

,

the number

is a triangular number if and only if

is a triangular number. (The triangular numbers are the

with in

.)


5/109

The Fiftieth Annual William Lowell Putnam Competition


A1 How many primes among the positive integers, writtenas usual in base 10, are alternating 1s and 0s, begin-

ning and ending with 1?

A2 Evaluate

where and

are positive.

A3 Prove that if

then

(Here

is a complex number and

.)

A4 If is an irrational number,

, is there afinite game with an honest coin such that the probabil-

ity of one player winning the game is ? (An honest

coin is one for which the probability of heads and the

probability of tails are both

. A game is finite if with

probability 1 it must end in a finite number of moves.)

A5 Let be a positive integer and let

be a regular

-gon inscribed in the unit circle. Show that

there is a positive constant , independent of

, with

the following property. For any points inside

there

are two distinct vertices

and

of such that

Here

denotes the distance between the points

and

.

A6 Let

be a formal power series

with coefficients in the field of two elements. Let

if every block of zeros in the binary

expansion of has an even number

of zeros in the block

otherwise.

(For example,

because

and

because

) Prove that

B1 A dart, thrown at random, hits a square target. Assum-

ing that any two parts of the target of equal area are

equally likely to be hit, find the probability that the point

hit is nearer to the center than to any edge. Express your

answer in the form

, where

are inte-

gers.

B2 Let be a non-empty set with an associative opera-tion that is left and right cancellative (

implies

, and

implies

). Assume that for

every in

the set

is finite.

Must be a group?

B3 Let be a function on

, differentiable and satis-

fying

for

. Assume that

for

(so

that

tends rapidly to

as

increases). For a

non-negative integer, define

(sometimes called the th moment of

).

a) Express in terms of .

b) Prove that the sequence

always con-

verges, and that the limit is

only if

.

B4 Can a countably infinite set have an uncountable collec-

tion of non-empty subsets such that the intersection of

any two of them is finite?

B5 Label the vertices of a trapezoid (quadrilateral

with two parallel sides) inscribed in the unit circle

as

so that

is parallel to

and

are in counterclockwise order. Let

,

and

denote the lengths of the line segments

,

and , where E is the point of intersection of the di-

agonals of , and is the center of the circle. Deter-

mine the least upper bound of

over all such for

which

, and describe all cases, if any, in which it

is attained.

B6 Let

be a point chosen at random from

the -dimensional region defined by

Let be a continuous function on

with

. Set

and

. Show that

the expected value of the Riemann sum

is

, where

is a polynomial of degree ,

independent of , with

for

.


6/109

The 51st William Lowell Putnam Mathematical Competition


A1 Let

T0= 2, T1= 3, T2= 6,

and forn3,Tn= (n + 4)Tn14nTn2+ (4n 8)Tn3.

The first few terms are

2, 3, 6, 14, 40, 152, 784, 5168, 40576.

Find, with proof, a formula for Tn of the form Tn =An+ Bn, where{An} and{Bn} are well-known se-quences.

A2 Is

2 the limit of a sequence of numbers of the form3

n 3m(n, m= 0, 1, 2, . . . )?A3 Prove that any convex pentagon whose vertices (no

three of which are collinear) have integer coordinates

must have area greater than or equal to 5/2.

A4 Consider a paper punch that can be centered at any pointof the plane and that, when operated, removes from the

plane precisely those points whose distance from the

center is irrational. How many punches are needed to

remove every point?

A5 IfA and B are square matrices of the same size suchthat ABAB = 0, does it follow that BABA = 0?

A6 If X is a finite set, let X denote the number of ele-ments inX. Call an ordered pair(S, T) of subsets of{1, 2, . . . , n} admissibleifs >|T| for eachsS, andt >|S|for eachtT. How many admissible orderedpairs of subsets of{1, 2, . . . , 10} are there? Prove youranswer.

B1 Find all real-valued continuously differentiable func-tionsfon the real line such that for allx,

(f(x))2 = x

0

[(f(t))2 + (f(t))2] dt + 1990.

B2 Prove that for |x|< 1, |z|> 1,

1 +

j=1

(1 + xj)Pj = 0,

wherePj is

(1 z)(1 zx)(1 zx2) (1 zxj1)(z x)(z x2)(z x3) (z xj) .

B3 Let S be a set of 2 2 integer matrices whose entries aij (1) are all squares of integers and, (2) satisfy aij 200. Show that ifShas more than 50387(= 154 152 15 + 2) elements, then it has two elements that commute.

B4 LetGbe a finite group of orderngenerated byaandbProve or disprove: there is a sequence

g1, g2, g3, . . . , g2n

such that

(1) every element ofG occurs exactly twice, and

(2) gi+1equalsgiaorgibfori= 1, 2, . . . , 2n. (Interpretg2n+1as g1.)

B5 Is there an infinite sequencea0, a1, a2, . . . of nonzerreal numbers such that for

n = 1, 2, 3, . . . the polyno

mial

pn(x) = a0+ a1x + a2x2 + + anxn

has exactlyn distinct real roots?

B6 LetSbe a nonempty closed bounded convex set in thplane. LetK be a line and t a positive number. LeL1 and L2 be support lines for Sparallel to K1, anletL be the line parallel toKand midway betweenLandL2. LetBS(K, t)be the band of points whose distance fromLis at most(t/2)w, wherew is the distanc

betweenL1and L2. What is the smallestt such that

SK

BS(K, t)=

for allS? (Kruns over all lines in the plane.)


7/109

The Fifty-Second William Lowell Putnam Mathematical Competition


A1 A

rectangle has vertices as

and

. It rotates

clockwise about the point

. It

then rotates


, then


, and finally,


. (The side originally

on the -axis is now back on the

-axis.) Find the

area of the region above the -axis and below the curve

traced out by the point whose initial position is (1,1).

A2 Let and

be different

matrices with real en-

tries. If and , can be

invertible?

A3 Find all real polynomials

of degree

for

which there exist real numbers

such that

1.

and

2.

where

denotes the derivative of

.

A4 Does there exist an infinite sequence of closed

discs

in the plane, with centers

, respectively, such that

1. the have no limit point in the finite plane,

2. the sum of the areas of the

is finite, and3. every line in the plane intersects at least one of the

?

A5 Find the maximum value of

for .

A6 Let denote the number of sums of positive inte-

gers

which add up to with

Let

denote the number of

which

add up to , with

1.

2. each is in the sequence

de-fined by

,

, and

and

3. if

then every element in

appears at least once as a

.

Prove that

for each

.

(For example, because the relevant sums are

and

because

the relevant sums are

)

B1 For each integer

, let

, where

is the greatest integer with

. Define a sequence

by

and

for

.

For what positive integers is this sequence eventually

constant?

B2 Suppose and

are non-constant, differentiable, real-

valued functions defined on

. Furthermore,

suppose that for each pair of real numbers and

,

If

, prove that

for all

.

B3 Does there exist a real number such that, if and

are integers greater than , then an

rectangle

may be expressed as a union of

and

rect-

angles, any two of which intersect at most along their

boundaries?

B4 Suppose is an odd prime. Prove that

B5 Let be an odd prime and let

denote (the field of)

integers modulo . How many elements are in the set

B6 Let

and be positive numbers. Find the largest num-

ber , in terms of

and , such that

for all

with

and for all ,

.

(Note:

.)


8/109

The 53rd William Lowell Putnam Mathematical Competition


A-1 Prove thatf(n) = 1nis the only integer-valued func-tion defined on the integers that satisfies the following

conditions.

(i) f(f(n)) = n, for all integersn;

(ii) f(f(n+ 2) + 2) = n for all integersn;

(iii) f(0) = 1.

A-2 DefineC()to be the coefficient ofx1992 in the powerseries aboutx = 0of(1 +x). Evaluate

10

C(y 1)

1992k=1

1

y+k

dy.

A-3 For a given positive integerm, find all triples (n,x,y)of positive integers, withnrelatively prime tom, whichsatisfy

(x2 +y2)m = (xy)n.

A-4 Letfbe an infinitely differentiable real-valued functiondefined on the real numbers. If

f

1

n

=

n2

n2 + 1, n= 1, 2, 3, . . . ,

compute the values of the derivatives f(k)(0), k =1, 2, 3, . . . .

A-5 For each positive integern, letan= 0(or 1) if the num-ber of 1s in the binary representation ofn is even (orodd), respectively. Show that there do not exist positiveintegersk and msuch that

ak+j =ak+m+j =ak+2m+j ,

for0 j m 1.

A-6 Four points are chosen at random on the surface of a

sphere. What is the probability that the center of the

sphere lies inside the tetrahedron whose vertices are atthe four points? (It is understood that each point is in-

dependently chosen relative to a uniform distribution on

the sphere.)

B-1 Let S be a set ofn distinct real numbers. LetAS bethe set of numbers that occur as averages of two distinct

elements ofS. For a givenn 2, what is the smallestpossible number of elements inAS?

B-2 For nonnegative integers n and k , defineQ(n, k)to bethe coefficient ofxk in the expansion of(1 +x+x2 +x3)n. Prove that

Q(n, k) =k

j=0

n

j

n

k 2j

,

whereab

is the standard binomial coefficient. (Re

minder: For integersaandbwitha 0,

ab

= a!

b!(ab)

for0 b a, with ab= 0otherwise.)

B-3 For any pair (x, y) of real numbers, a sequenc(an(x, y))n0 is defined as follows:

a0(x, y) = x,

an+1(x, y) = (an(x, y))

2 +y2

2 , forn 0.

Find the area of the region

{(x, y)|(an(x, y))n0 converges}.

B-4 Letp(x) be a nonzero polynomial of degree less tha1992 having no nonconstant factor in common with

x3 x. Let

d1992

dx1992

p(x)

x3 x

=

f(x)

g(x)

for polynomialsf(x) and g(x). Find the smallest possible degree off(x).

B-5 LetDndenote the value of the(n 1) (n 1)determinant

3 1 1 1 11 4 1 1 11 1 5 1 11 1 1 6 1...

......

.... . .

...

1 1 1 1 n+ 1

.

Is the setDnn!

n2

bounded?

B-6 Let M be a set of realn nmatrices such that

(i) I M, whereIis then nidentity matrix;

(ii) ifA Mand B M, then either AB MoAB M, but not both;

(iii) ifA M andB M, then eitherAB = BAoAB= BA;

(iv) ifA M andA=I, there is at least oneB Msuch thatAB = BA.

Prove that M contains at mostn2 matrices.


9/109

The 54th William Lowell Putnam Mathematical Competition


A1 The horizontal liney = c intersects the curvey = 2x3x3 in the first quadrant as in the figure. Findc so thatthe areas of the two shaded regions are equal. [Figure

not included. The first region is bounded by they-axis,the line y = c and the curve; the other lies under thecurve and above the liney = cbetween their two pointsof intersection.]

A2 Let (xn)n0 be a sequence of nonzero real numberssuch that x2n xn1xn+1 = 1 for n = 1, 2, 3, . . . .Prove there exists a real number a such that xn+1 =axn xn1 for alln 1.

A3 Let Pn be the set of subsets of {1, 2, . . . , n}. Letc(n, m) be the number of functions f : Pn {1, 2, . . . , m} such that f(AB) = min{f(A), f(B)}.Prove that

c(n, m) =mj=1

jn.

A4 Let x1, x2, . . . , x19be positive integers each of which isless than or equal to 93. Lety1, y2, . . . , y93 be positiveintegers each of which is less than or equal to 19. Prove

that there exists a (nonempty) sum of somexis equalto a sum of someyjs.

A5 Show that

10100

x2 xx3 3x+ 1

2dx+

111

1

101

x2 x

x3 3x+ 1

2dx+

1110

101

100

x2 x

x3 3x+ 1

2dx

is a rational number.

A6 The infinite sequence of 2s and 3s

2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3,

3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 2, . . .

has the property that, if one forms a second sequencethat records the number of 3s between successive 2s,

the result is identical to the given sequence. Show that

there exists a real numberrsuch that, for anyn, thenthterm of the sequence is 2 if and only ifn = 1 + rmfor some nonnegative integer m. (Note:x denotes thelargest integer less than or equal tox.)

B1 Find the smallest positive integern such that for everintegermwith0< m < 1993, there exists an integerkfor which

m

1993 N.B4 Forn 1, letdn be the greatest common divisor of th

entries ofAn I, where

A=

3 24 3

and I=

1 00 1

.

Show thatlimn dn= .

B5 For any real number , define the function f(x) =x. Letnbe a positive integer. Show that there existan such that for1 k n,

fk(n2) =n2 k= fk(n2).

B6 For any integern, set

na= 101a 100 2a.

Show that for0 a,b,c,d 99,na+ nb nc+ n(mod 10100)implies {a, b} = {c, d}.


11/109

The Fifty-Sixth William Lowell Putnam Mathematical Competition


A1 Let

be a set of real numbers which is closed under

multiplication (that is, if and

are in

, then so is ).

Let and be disjoint subsets of

whose union is

. Given that the product of any three(not necessarilydistinct) elements of is in and that the product of

any three elements of is in

, show that at least one

of the two subsets

is closed under multiplication.

A2 For what pairs of positive real numbers does the

improper integral

converge?

A3 The number has nine (not necessarily dis-

tinct) decimal digits. The number

is such

that each of the nine 9-digit numbers formed by replac-ing just one of the digits

is

by the corre-

sponding digit (

) is divisible by 7. The

number

is related to

is the same

way: that is, each of the nine numbers formed by replac-

ing one of the

by the corresponding

is divisible by

7. Show that, for each ,

is divisible by 7. [For

example, if

, then may be 2

or 9, since

and

are multiples of

7.]

A4 Suppose we have a necklace of beads. Each bead is

labeled with an integer and the sum of all these labels

is

. Prove that we can cut the necklace to form a

string whose consecutive labels

satisfy

for

A5 Let

be differentiable (real-valued) func-

tions of a single variable which satisfy

..

.

..

.

for some constants

. Suppose that for all ,

as

. Are the functions

necessarily linearly dependent?

A6 Suppose that each of people writes down the numbers

1,2,3 in random order in one column of a

matrix,

with all orders equally likely and with the orders for

different columns independent of each other. Let the

row sums

of the resulting matrix be rearranged

(if necessary) so that . Show that for some

, it is at least four times as likely that both

and

as that .

B1 For a partition of

, let

be

the number of elements in the part containing

. Prove

that for any two partitions and

, there are two dis-

tinct numbers

and in such

that

and

. [A partition of

a set

is a collection of disjoint subsets (parts) whose

union is

.]

B2 An ellipse, whose semi-axes have lengths and , rolls

without slipping on the curve . How are

related, given that the ellipse completes one rev-

olution when it traverses one period of the curve?

B3 To each positive integer with

decimal digits, we as-

sociate the determinant of the matrix obtained by writ-

ing the digits in order across the rows. For example, for

, to the integer 8617 we associate

. Find, as a function of , the sum of all the determi-

nants associated with

-digit integers. (Leading digits

are assumed to be nonzero; for example, for ,

there are 9000 determinants.)

B4 Evaluate

Express your answer in the form

, where

are integers.

B5 A game starts with four heaps of beans, containing 3,4,5

and 6 beans. The two players move alternately. A move

consists of takingeither

a) one bean from a heap, provided at least two beans

are left behind in that heap, or

b) a complete heap of two or three beans.

The player who takes the last heap wins. To win thegame, do you want to move first or second? Give a

winning strategy.

B6 For a positive real number , define

Prove that cannot be expressed as the dis-

joint union of three sets

and

. [As

usual,

is the greatest integer

.]


12/109

Solutions to the Fifty-Sixth William Lowell Putnam Mathematical Competition


Kiran Kedlaya

A1 Suppose on the contrary that there exist

with

and

with

. Then

while

, contradiction.

A2 The integral converges iff . The easiest proof

uses big-O notation and the fact that

for

. (Here

means

bounded by a constant times

.)

So

hence

and similarly

Hence the integral were looking at is

The term

is bounded by a constant times

, whose integral converges. Thus we only have todecide whether

converges. But

has divergent integral, so we get convergence if and

only if (in which case the integral telescopes any-

way).

A3 Let and

be the numbers

and

, re-

spectively. We are given that

and

for

. Sum the first relation over

and we

get

, or

.

Now add the first and second relations for any particu-

lar value of and we get

. But we know

is divisible by 7, and 10

is coprime to 7, so

.

A4 Let

, so that

. These form a cyclic sequence that doesnt change

when you rotate the necklace, except that the entire se-

quence gets translated by a constant. In particular, it

makes sense to choose

for which is maximum and

make that one

; this way

for all , which gives

, but the right side may be

replaced by

since the left side is an integer.

A5 Everyone (presumably) knows that the set of solutions

of a system of linear first-order differential equations

with constant coefficients is

-dimensional, with ba-

sis vectors of the form

(i.e. a function times

a constant vector), where the

are linearly indepen-

dent. In particular, our solution

can be written as

.

Choose a vector

orthogonal to

but not to

. Since

as

, the same is true of

;

but that is simply

. In other words, if

, then

must also go to 0.

However, it is easy to exhibit a solution which does

not go to 0. The sum of the eigenvalues of the matrix

, also known as the trace of

, being the sum

of the diagonal entries of

, is nonnegative, so

has

an eigenvalue with nonnegative real part, and a cor-

responding eigenvector

. Then

is a solution that

does not go to 0. (If is not real, add this solution to

its complex conjugate to get a real solution, which still

doesnt go to 0.)

Hence one of the

, say

, is zero, in which case

for all

.

A6 View this as a random walk/Markov process with states

the triples of integers with sum 0, correspond-

ing to the difference between the first, second and third

rows with their average (twice the number of columns).

Adding a new column adds on a random permutation

of the vector

. I prefer to identify the triple

with the point

in

the plane, where

is a cube root of unity. Then adding

a new column corresponds to moving to one of the six

neighbors of the current position in a triangular lattice.

What wed like to argue is that for large enough

, the

ratio of the probabilities of being in any two particular

states goes to 1. Then in fact, well see that eventually,about six times as many matrices have

than

. This is a pain to prove, though,

and in fact is way more than we actually need.

Let and

be the probability that we are at the

origin, or at a particular point adjacent to the origin,

respectively. Then

. (In fact,

is

times the sum of the probabilities of being at each

neighbor of the origin at time

, but these are all

.)


13/109

So the desired result, which is that

for

some large

, is equivalent to

.

Suppose on the contrary that this is not the case; then

for some constant

. However, if

, the probability that we chose each of the six

types of moves

times is already

,

which by Stirlings approximation is asymptotic to a

constant times

. This term alone is bigger than

, so we must have

for some

. (In fact, we must have

for any

.)

B1 For a given , no more than three different values of

are possible (four would require one part each of

size at least 1,2,3,4, and thats already more than 9 el-

ements). If no such

exist, each pair

occurs for at most 1 element of

, and since there are

only

possible pairs, each must occur exactly once.

In particular, each value of

must occur 3 times.

However, clearly any given value of

occurs

times, where is the number of distinct partitions of

that size. Thus

can occur 3 times only if it equals

1 or 3, but we have three distinct values for which it

occurs, contradiction.

B2 For those who havent taken enough physics, rolling

without slipping means that the perimeter of the ellipse

and the curve pass at the same rate, so all were saying

is that the perimeter of the ellipse equals the length of

one period of the sine curve. So set up the integrals:

Let

in the second integral and write 1 as

and you get

Since the left side is increasing as a function of , we

have equality if and only if

.

B3 For

we obviously get 45, while for

the

answer is 0 because it both changes sign (because de-terminants are alternating) and remains unchanged (by

symmetry) when you switch any two rows other than

the first one. So only

is left. By the multilin-

earity of the determinant, the answer is the determinant

of the matrix whose first (resp. second) row is the sum

of all possible first (resp. second) rows. There are 90

first rows whose sum is the vector

, and 100

second rows whose sum is

. Thus the answer

is

B4 The infinite continued fraction is defined as the limit

of the sequence

.

Notice that the sequence is strictly decreasing (by in-

duction) and thus indeed has a limit , which satisfies

, or rewriting,

.

Moreover, we want the greater of the two roots.

Now how to compute the eighth root of ? Notice that

if

satisfies the quadratic

, then we

have

Clearly, then, the positive square roots of the quadratic

satisfy the quadratic

. Thus we compute that

is the greater

root of

,

is the greater root

of

, and

is the greater root of

, otherwise known as

.

B5 This problem is dumb if you know the Sprague-

Grundy theory of normal impartial games (see Conway,Berlekamp and Guy,Winning Ways, for details). Ill de-

scribe how it applies here. To each position you assign

a nim-value as follows. A position with no moves (in

which case the person to move has just lost) takes value

0. Any other position is assigned the smallest number

not assigned to a valid move from that position.

For a single pile, one sees that an empty pile has value

0, a pile of 2 has value 1, a pile of 3 has value 2, a pile

of 4 has value 0, a pile of 5 has value 1, and a pile of 6

has value 0.

You add piles just like in standard Nim: the nim-value

of the composite of two games (where at every turn you

pick a game and make a move there) is the base 2 ad-dition without carries (i.e. exclusive OR) of the nim-

values of the constituents. So our starting position, with

piles of 3, 4, 5, 6, has nim-value

.

A position is a win for the player to move if and only if

it has a nonzero value, in which case the winning strat-

egy is to always move to a 0 position. (This is always

possible from a nonzero position and never from a zero

position, which is precisely the condition that defines

the set of winning positions.) In this case, the winning

move is to reduce the pile of 3 down to 2, and you can

easily describe the entire strategy if you so desire.

B6 Obviously

have to be greater than 1, and no twocan both be rational, so without loss of generality as-

sume that and

are irrational. Let

denote the fractional part of

. Then

if and

only if

. In particular, this

means that

contains

elements, and similarly. Hence for every integer

,

2


14/109

Dividing through by

and taking the limit as

shows that

. That in turn implies

that for all

,

Our desired contradiction is equivalent to showing that

the left side actually takes the value 1 for some

.

Since the left side is an integer, it suffices to show that

for some

.

A result in ergodic theory (the two-dimensional version

of the Weil equidistribution theorem) states that if

are linearly independent over the rationals, then the set

of points

is dense (and in fact equidis-

tributed) in the unit square. In particular, our claim def-

initely holds unless

for some integers

.

On the other hand, suppose that such a relation

does hold. Since and

are irrational, by the

one-dimensional Weil theorem, the set of points

is dense in the set of

in the

unit square such that

is an integer. It is simpleenough to show that this set meets the region

unless

is an integer, and that

would imply that

, a quantity between 0 and

1, is an integer. We have our desired contradiction.

3


15/109

The Fifty-Seventh William Lowell Putnam Mathematical Competition


A1 Find the least number

such that for any two squares of

combined area 1, a rectangle of area

exists such that

the two squares can be packed in the rectangle (without

interior overlap). You may assume that the sides of thesquares are parallel to the sides of the rectangle.

A2 Let

and

be circles whose centers are 10 units

apart, and whose radii are 1 and 3. Find, with proof, the

locus of all points

for which there exists points on

and

on

such that

is the midpoint of the line

segment

.

A3 Suppose that each of 20 students has made a choice of

anywhere from 0 to 6 courses from a total of 6 courses

offered. Prove or disprove: there are 5 students and 2

courses such that all 5 have chosen both courses or all 5

have chosen neither course.

A4 Let be the set of ordered triples

of distinct

elements of a finite set

. Suppose that

1.

if and only if

;

2.

if and only if

;

3.

and

are both in if and only if

and

are both in

.

Prove that there exists a one-to-one function from

to

such that implies .

Note:

is the set of real numbers.

A5 If is a prime number greater than 3 and

,

prove that the sum

of binomial coefficients is divisible by

.

A6 Let

be a constant. Give a complete descrip-

tion, with proof, of the set of all continuous functions

such that

for all

.

Note that

denotes the set of real numbers.

B1 Define a selfishset to be a set which has its own cardi-

nality (number of elements) as an element. Find, with

proof, the number of subsets of

which are

minimalselfish sets, that is, selfish sets none of whose

proper subsets is selfish.

B2 Show that for every positive integer

,

B3 Given that

, find,

with proof, the largest possible value, as a function of

(with ), of

B4 For any square matrix

, we can define

by the

usual power series:

Prove or disprove: there exists a

matrix

with

real entries such that

B5 Given a finite string of symbols and , we write

for the number of s in minus the number

of s. For example,

. We

call a string balanced if every substring of (con-

secutive symbols of) has

. Thus,

is not balanced, since it contains the sub-

string

. Find, with proof, the number of bal-

anced strings of length .

B6 Let

be the vertices of a

convex polygon which contains the origin in its inte-

rior. Prove that there exist positive real numbers and

such that


16/109

Solutions to the Fifty-Eighth William Lowell Putnam Mathematical Competition


Manjul Bhargava and Kiran Kedlaya

A-1 If and are the sides of two squares with combined

area 1, then

. Suppose without loss of gen-

erality that

. Then the shorter side of a rectangle

containing both squares without overlap must be at least , and the longer side must be at least

. Hence the

desired value of is the maximum of

.

To find this maximum, we let

with . Then we are to maximize

with equality for . Hence this value is the de-

sired value of .

A-2 Let

and

be the centers of

and

, respec-

tively. (We are assuming

has radius 1 and

has

radius 3.) Then the desired locus is an annulus centered

at the midpoint of

, with inner radius 1 and outer

radius 2.

For a fixed point on

, the locus of the midpoints of

the segments for lying on is the image of

under a homothety centered at of radius

, which

is a circle of radius

. As varies, the center of this

smaller circle traces out a circle

of radius

(again

by homothety). By considering the two positions of

on the line of centers of the circles, one sees that

is

centered at the midpoint of

, and the locus is now

clearly the specified annulus.

A-3 The claim is false. There are

ways to choose

3 of the 6 courses; have each student choose a different

set of 3 courses. Then each pair of courses is chosen by

4 students (corresponding to the four ways to complete

this pair to a set of 3 courses) and is not chosen by 4

students (corresponding to the 3-element subsets of the

remaining 4 courses).

Note: Assuming that no two students choose the same

courses, the above counterexample is unique (up to per-

muting students). This may be seen as follows: Given a

group of students, suppose that for any pair of courses

(among the six) there are at most 4 students taking both,

and at most 4 taking neither. Then there are at most

pairs

, where is a student, and

is a set of two courses of which is taking either both

or none. On the other hand, if a student is taking

courses, then he/she occurs in

such

pairs

. As

is minimized for

, it follows

that every student occurs in at least

such

pairs

. Hence there can be at most

stu-

dents, with equality only if each student takes 3 courses,

and for each set of two courses, there are exactly 4 stu-

dents who take both and exactly 4 who take neither.

Since there are only 4 ways to complete a given pair

of courses to a set of 3, and only 4 ways to choose 3

courses not containing the given pair, the only way for

there to be 20 students (under our hypotheses) is if allsets of 3 courses are in fact taken. This is the desired

conclusion.

However, Robin Chapman has pointed out that the so-

lution is not unique in the problem as stated, because a

given selection of courses may be made by more than

one student. One alternate solution is to identify the 6

courses with pairs of antipodal vertices of an icosahe-

dron, and have each student pick a different face and

choose the three vertices touching that face. In this ex-

ample, each of 10 selections is made by a pair of stu-

dents.

A-4 In fact, we will show that such a function

exists withthe property that

if and only if

for some cyclic permutation

of

. We proceed by induction on the number of el-

ements in . If and , then

choose with , otherwise choose

with .

Now let be an element of

and

.

Let be the elements of labeled such that

. We claim that there ex-

ists a unique such that ,

where hereafter

.

We show existence first. Suppose no such exists; then

for all

, we have

.

This holds by property 1 for and by induction on

in general, noting that

Applying this when

, we get

,


17/109

contradicting the fact that

. Hence ex-

istence follows.

Now we show uniqueness. Suppose

; then for any , we have

by the assumption on

. Therefore

so

. The case

is ruled out by

and the case is similar.

Finally, we put in if , and

otherwise; an analysis similar to that

above shows that has the desired property.

A-5 (due to Lenny Ng) For

, divides

and

where the congruence

means that

is a rational number whose numerator, in reduced form,

is divisible by . Hence it suffices to show that

We distinguish two cases based on . First

suppose

, so that

. Then

since

.

Now suppose

, so that

. A similar

argument gives

A-6 We first consider the case

; we shall show in

this case must be constant. The relation

proves that is an even function. Let

be the

roots of , both of which are real. If

,

define

and

for each positiveinteger . By induction on ,

for all , so the sequence tends to a limit

which is a

root of not less than

. Of course this means

. Since for all and

,

we conclude

, so is constant on

.

If

and is defined as before, then by in-

duction,

. Note that the sequence can

be defined because

; the latter follows by noting

that the polynomial is positive at and

has its minimum at

, so both roots are greater

than . In any case, we deduce that is also constant

on

.

Finally, suppose

. Now define

. Given that

, we have

. Thus

if we had

for all , by the same argument as in

the first case we deduce

and so

.

Actually, this doesnt happen; eventually we have

, in which case by what we

have already shown. We conclude that is a constant

function. (Thanks to Marshall Buck for catching an in-

accuracy in a previous version of this solution.)

Now suppose

. Then the sequence defined

by and

is strictly increasing

and has no limit point. Thus if we define on

as any continuous function with equal values on the

endpoints, and extend the definition from

to

by the relation

, and

extend the definition further to

by the relation

, the resulting function has the desired

property. Moreover, any function with that property

clearly has this form.

B-1 Let denote the set

, and let denote

the number of minimal selfish subsets of . Then the

number of minimal selfish subsets of not containing

2


18/109

is equal to

. On the other hand, for any mini-

mal selfish subset of containing , by subtracting 1

from each element, and then taking away the element from the set, we obtain a minimal selfish subset

of

(since and cannot both occur in a selfish

set). Conversely, any minimal selfish subset of

gives rise to a minimal selfish subset of containing by the inverse procedure. Hence the number of min-

imal selfish subsets of

containing

is

. Thuswe obtain

. Since

, we

have , where denotes the th term of the

Fibonacci sequence.

B-2 By estimating the area under the graph of using up-

per and lower rectangles of width 2, we get

Since

, we have, upon expo-

nentiating and taking square roots,

using the fact that

.

B-3 View

as an arrangement of the numbers

on a circle. We prove that the optimal ar-

rangement is

To show this, note that if is a pair of adjacent num-

bers and is another pair (read in the same order

around the circle) with and , then the seg-

ment from to can be reversed, increasing the sum

by

Now relabel the numbers so they appear in order as fol-lows:

where without loss of generality we assume

. By considering the pairs

and

and using the trivial fact

, we

deduce

. We then compare the pairs

and

, and using that

, we deduce

. Continuing in this

fashion, we prove that

and

so

for

, i.e. that the optimal ar-

rangement is as claimed. In particular, the maximum

value of the sum is

Alternate solution: We prove by induction that the value

given above is an upper bound; it is clearly a lower

bound because of the arrangement given above. As-

sume this is the case for . The optimal arrangement

for is obtained from some arrangement for by

inserting between some pair of adjacent terms.

This operation increases the sum by

, which is an increasing function of

both and . In particular, this difference is maximal

when and equal and

. Fortunately, this

yields precisely the difference between the claimed up-

per bound for and the assumed upper bound for ,

completing the induction.

B-4 Suppose such a matrix exists. If the eigenvalues of

(over the complex numbers) are distinct, then there

exists a complex matrix such that

is

diagonal. Consequently,

is diagonal. But then

must be diagonalizable, a con-

tradiction. Hence the eigenvalues of are the same,

and has a conjugate

over the complex

numbers of the form

A direct computation shows that

Since

and

are conjugate, their eigenvalues

must be the same, and so we must have . This

implies , so that is the identity matrix,

as must be , a contradiction. Thus cannot exist.

Alternate solution (due to Craig Helfgott and AlexPopa): Define both

and

by the usual power

series. Since commutes with itself, the power series

identity

holds. But if is the given matrix, then by the above

identity,

must equal

which is

3


19/109

a nilpotent matrix. Thus is also nilpotent. How-

ever, the square of any

nilpotent matrix must be

zero (e.g., by the Cayley-Hamilton theorem). This is a

contradiction.

B-5 Consider a

checkerboard, in which we write an -letter string, one letter per square. If the string is

balanced, we can cover each pair of adjacent squares

containing the same letter with a

domino, and

these will not overlap (because no three in a row can

be the same). Moreover, any domino is separated from

the next by an even number of squares, since they must

cover opposite letters, and the sequence must alternate

in between.

Conversely, any arrangement of dominoes where ad-

jacent dominoes are separated by an even number of

squares corresponds to a unique balanced string, once

we choose whether the string starts with or . In

other words, the number of balanced strings is twice

the number of acceptable domino arrangements.

We count these arrangements by numbering the squares and distinguishing whether the dominoes

start on even or odd numbers. Once this is decided, one

simply chooses whether or not to put a domino in each

eligible position. Thus we have

arrangements in

the first case and

in the second, but note that

the case of no dominoes has been counted twice. Hence

the number of balanced strings is

B-6 We will prove the claim assuming only that the convex

hull of the points contains the origin in its in-

terior. (Thanks to Marshall Buck for pointing out that

the last three words are necessary in the previous sen-

tence!) Let

so that the left-hand

side of the given equation is

(1)

Now note that (1) is the gradient of the function

and so it suffices to show has a critical point. We will

in fact show has a global minimum.

Clearly we have

Note that this maximum is positive for :

if we had for all , then the subset

of the -plane would be a half-plane

containing all of the points , whose convex hull

would then not contain the origin, a contradiction.

The function

is clearly continuous on

the unit circle

, which is compact. Hence it

has a global minimum , and so for all ,

In particular,

on the disk of radius

. Since

, the infimum of

is the same over the entire -plane as over this disk,

which again is compact. Hence attains its infimal

value at some point in the disk, which is the desired

global minimum.

Noam Elkies has suggested an alternate solution as fol-

lows: for , draw the loop traced by (1) as

travels counterclockwise around the circle .

For , this of course has winding number 0 about

any point, but for large, one can show this loop has

winding number 1 about the origin, so somewhere in

between the loop must pass through the origin. (Prov-

ing this latter fact is a little tricky.)

4


20/109

The Fifty-Eighth William Lowell Putnam Mathematical Competition


A1 A rectangle,

, has sides

and

. A triangle

has

as the intersection of the al-

titudes,

the center of the circumscribed circle,

the

midpoint of

, and

the foot of the altitude from .

What is the length of

?

A2 Players

are seated around a table, and

each has a single penny. Player 1 passes a penny to

player 2, who then passes two pennies to player 3.

Player 3 then passes one penny to Player 4, who passes

two pennies to Player 5, and so on, players alternately

passing one penny or two to the next player who still

has some pennies. A player who runs out of pennies

drops out of the game and leaves the table. Find an in-

finite set of numbers

for which some player ends upwith all

pennies.

A3 Evaluate

A4 Let be a group with identity and a

function such that

whenever

. Prove that thereexists an element such that

is a homomorphism (i.e.

for all

).

A5 Let denote the number of ordered

-tuples of posi-

tive integers

such that

. Determine whether

is even or odd.

A6 For a positive integer

and any real number , define

recursively by

,

, and for

,

Fix

and then take to be the largest value for which

. Find

in terms of

and ,

.

B1 Let

denote the distance between the real number

and the nearest integer. For each positive integer

,

evaluate

(Here

denotes the minimum of and

.)

B2 Let be a twice-differentiable real-valued function sat-

isfying

where

for all real

. Prove that

is

bounded.

B3 For each positive integer

, write the sum

in the form

, where

and

are relatively prime

positive integers. Determine all

such that 5 does not

divide

.

B4 Let

denote the coefficient of

in the expansion

of

. Prove that for all [integers]

,

B5 Prove that for

,

terms

terms

B6 The dissection of the 345 triangle shown below (into

four congruent right triangles similar to the original) has

diameter

. Find the least diameter of a dissection of

this triangle into four parts. (The diameter of a dissec-

tion is the least upper bound of the distances betweenpairs of points belonging to the same part.)


21/109

Solutions to the Fifty-Eighth William Lowell Putnam Mathematical Competition


Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 The centroid

of the triangle is collinear with and

(Euler line), and the centroid lies two-thirds of the

way from to

. Therefore

is also two-thirds of

the way from to

, so

. Since the triangles

and

are similar (theyre right triangles and

), we have

, or

. Now

, but

, so

A2 We show more precisely that the game terminates with

one player holding all of the pennies if and only if

or

for some

. First sup-

pose we are in the following situation for some

.

(Note: for us, a move consists of two turns, starting

with a one-penny pass.)

Except for the player to move, each player has

pennies;

The player to move has at least pennies.

We claim then that the game terminates if and only if

the number of players is a power of 2. First supposethe number of players is even; then after complete

rounds, every other player, starting with the player who

moved first, will have more pennies than initially, and

the others will all have 0. Thus we are reduced to the

situation with half as many players; by this process, we

eventually reduce to the case where the number of play-

ers is odd. However, if there is more than one player,

after two complete rounds everyone has as many pen-

nies as they did before (here we need

), so the

game fails to terminate. This verifies the claim.

Returning to the original game, note that after one com-

plete round,

players remain, each with 2 pennies

except for the player to move, who has either 3 or 4 pen-nies. Thus by the above argument, the game terminates

if and only if

is a power of 2, that is, if and only

if

or

for some .

A3 Note that the series on the left is simply

.

By integration by parts,

and so by induction,

Thus the desired integral is simply

A4 In order to have for all , we must in par-

ticular have this for , and so we take

.

We first note that

and so

commutes with

for all

. Next, we

note that

and using the commutativity of , we deduce

or

, as desired.

A5 We may discard any solutions for which

, since

those come in pairs; so assume

. Similarly, wemay assume that

,

,

,

.

Thus we get the equation

Again, we may assume

and

, so we get

; and

, so

. This implies that

, which by

counting has 5 solutions. Thus

is odd.

A6 Clearly

is a polynomial in of degree , so it suf-

fices to identify values of for which

. We

claim these are

for

; in

this case,

is the coefficient of

in the polynomial

. This can be verified bynoticing that

satisfies the differential equation

(by logarithmic differentiation) or equivalently,


22/109

and then taking the coefficient of

on both sides:

In particular, the largest such is , and

for

.

Greg Kuperberg has suggested an alternate approach to

show directly that

is the largest root, withoutcomputing the others. Note that the condition

states that

is an eigenvector of the matrix

otherwise

with eigenvalue . By the Perron-Frobenius theorem,

has a unique eigenvector with positive entries, whose

eigenvalue has modulus greater than or equal to that of

any other eigenvalue, which proves the claim.

B1 It is trivial to check that

for

, that

for

, that

for

, and that

for

. Therefore the desired sum is

B2 It suffices to show that

is bounded for ,

since

satisfies the same equation as

. But

then

so that

for

.

B3 The only such are the numbers 14, 2024, 100104,

and 120124. For the proof let

and introduce the auxiliary function

It is immediate (e.g., by induction) that

(mod ) for

(mod 5) re-

spectively, and moreover, we have the equality

where denotes the largest integer such that

. We wish to determine those such that the

above sum has nonnegative 5valuation. (By the 5

valuation of a number we mean the largest integer

such that

is an integer.)

If

, then the last term in the above sum

has 5valuation , since

,

,

each have valu-

ation 0; on the other hand, all other terms must have

5valuation strictly larger than . It follows that

has 5valuation exactly ; in particular,

has non-

negative 5valuation in this case if and only if ,

i.e., , 2, or 3.

Suppose now that

. Then we must also

have

. The former condition im-

plies that the last term of the above sum is

, which has 5valuation

.

It is clear that

(mod 25); hence if

equals 20 or 24, then the secondtolast term

of the above sum (if it exists) has valuation at least

. The thirdtolast term (if it exists) is of

the form

, so that the sum of the last term andthe third to last term takes the form

.

Since

can be congruent only to 0,1, or -1 (mod 5),

and

(mod 5), we conclude that the sum of the

last term and thirdtolast term has valuation

,

while all other terms have valuation strictly higher.

Hence

has nonnegative5valuation in this case only

when

, leading to the values

(arising from

), 20,24 (arising from

and

and 24 resp.), 101, 102, 103, and 104 (arising from

,

) and 120, 121, 122, 123, and

124 (arising from ,

).

Finally, suppose

and

, 22,

or 23. Then as before, the first condition implies thatthe last term of the sum in (*) has valuation

,

while the second condition implies that the secondto

last term in the same sum has valuation

. Hence

all terms in the sum (*) have 5valuation strictly higher

than

, except for the secondtolast term, and

therefore

has 5valuation

in this case. In

particular,

is integral (mod 5) in this case if and only

if

, which gives the additional values

, 22,

and 23.

B4 Let

be the given sum (note that

is nonzero precisely for

. Since

we have

2


23/109

By computing

, we may eas-

ily verify by induction that

and

for all

. (Alternate so-

lution suggested by John Rickert: write

, and note note that is the

coefficient of

in

.)

B5 Define the sequence

,

for .

It suffices to show that for every

,

for some

. We do this by induction on , with

being obvious.

Write

, where is odd. It suffices to show

that

modulo

and modulo , for some

. For the former, we only need

,

but clearly

by induction on . For the lat-

ter, note that

as long as

, where is the Eu-

ler totient function. By hypothesis, this occurs for some

. (Thanks to Anoop Kulkarni for

catching a lethal typo in an earlier version.)

B6 The answer is

. Place the triangle on the carte-

sian plane so that its vertices are at

. It is easy to check that the five points

and

are

all in the triangle and have distance at least

apart

from each other (note that

); thus any dis-

section of the triangle into four parts must have diame-

ter at least

.

We now exhibit a dissection with least diameter

.(Some variations of this dissection are possible.) Put

,

,

,

, and divide

into

the convex polygonal regions

,

,

,

; each region has diameter

, as can be

verified by checking the distance between each pair of

vertices of each polygon. (One need only check for the

pentagon: note that and

are contained in

circular sectors centered at and

, respectively, of ra-

dius

and angle less than

, and that

is

a rectangle with diagonal

.)

3


24/109



A1 A right circular cone has base of radius 1 and height 3.A cube is inscribed in the cone so that one face of the

cube is contained in the base of the cone. What is the

side-length of the cube?

A2 Lets be any arc of the unit circle lying entirely in thefirst quadrant. LetA be the area of the region lyingbelows and above thex-axis and let B be the area ofthe region lying to the right of they-axis and to the leftofs. Prove thatA +B depends only on the arc length,and not on the position, ofs.

A3 Let fbe a real function on the real line with continuousthird derivative. Prove that there exists a pointa suchthat

f(a)

f(a)

f(a)

f(a)

0.

A4 Let A1 = 0 and A2 = 1. For n > 2, the num-berAn is defined by concatenating the decimal expan-sions ofAn1 and An2 from left to right. For ex-ample A3 = A2A1 = 10, A4 = A3A2 = 101,A5 = A4A3 = 10110, and so forth. Determine allnsuch that11 divides An.

A5 LetFbe a finite collection of open discs in R2 whoseunion contains a set E R2. Show that there is apairwise disjoint subcollection D1, . . . , Dn inF suchthat

E nj=13Dj.Here, ifD is the disc of radius rand centerP, then3Dis the disc of radius3rand centerP.

A6 LetA, B,Cdenote distinct points with integer coordi-nates in R2. Prove that if

(|AB| + |BC|)2 0.

B2 Given a point(a, b)with0< b < a, determine the minimum perimeter of a triangle with one vertex at (a, b)one on thex-axis, and one on the liney =x. You maassume that a triangle of minimum perimeter exists.

B3 let H be the unit hemisphere{(x,y,z) : x2 +y2 +z2 = 1, z 0}, C the unit circle{(x,y, 0) : x2 +y2 = 1}, and Pthe regular pentagon inscribed in CDetermine the surface area of that portion ofH lyin

over the planar region insideP, and write your answein the form A sin + B cos , where A,B,,are reanumbers.

B4 Find necessary and sufficient conditions on positive in

tegersmandnso that

mn1

i=0

(1)i/m+i/n = 0.

B5 LetN

be the positive integer with 1998 decimal digits

all of them 1; that is,

N= 1111 11.

Find the thousandthdigit after the decimal point of

N

B6 Prove that, for any integersa, b, c, there exists a positive integern such that

n3 + an2 + bn + cis not a

integer.


25/109

Solutions to the 59th William Lowell Putnam Mathematical Competition



A1 Consider the plane containing both the axis of the cone

and two opposite vertices of the cubes bottom face.

The cross section of the cone and the cube in this plane

consists of a rectangle of sides and inscribed in

an isosceles triangle of base and height , where is

the side-length of the cube. (The side of the rect-

angle lies on the base of the triangle.) Similar triangles

yield

, or

A2 First solution: to fix notation, let be the area of re-

gion

, and be the area of

; further let

denote the area of sector

, which only depends

on the arc length of

. If

denotes the area of tri-angle

, then we have

and

. But clearly

and

, and so

.

O

D

E

F G

H

I

Second solution: We may parametrize a point in by

any of ,

, or

. Then and

are

just the integrals of

and

over the appropriate

intervals; thus

is the integral of

(mi-

nus because the limits of integration are reversed). But

, and so

is precisely the

radian measure of . (Of course, one can perfectly well

do this problem by computing the two integrals sepa-

rately. But whats the fun in that?)

A-3 If at least one of

,

,

, or

van-

ishes at some point

, then we are done. Hence wemay assume each of , , , and

is either strictly positive or strictly negative on the

real line. By replacing by if necessary,

we may assume ; by replacing by

if necessary, we may assume . (No-

tice that these substitutions do not change the sign of

.) Now

implies that

is increasing, and

implies that

is convex, so that

for all

and

. By letting

increase in the latter inequality,

we see that

must be positive for sufficiently

large ; it follows that

for all . Similarly,

and

imply that

for all

. Therefore

for all , and

we are done.

A-4 The number of digits in the decimal expansion of

is the Fibonacci number , where

,

,

and

for

. It follows that

the sequence

, modulo 11, satisfies the recursion

. (Notice that the recur-

sion for

depends only on the value of

modulo2.) Using these recursions, we find that and

modulo 11, and that

and

mod-

ulo 2. It follows that

(mod 11) for all

. We find that among

, and

, only

vanishes modulo 11. Thus 11 divides

if and only if

for some nonnegative integer

.

A5 Define the sequence by the following greedy algo-

rithm: let

be the disc of largest radius (breaking ties

arbitrarily), let

be the disc of largest radius not meet-

ing

, let

be the disc of largest radius not meeting

or

, and so on, up to some final disc

. To see

that

, consider a point in ; if it lies inone of the

, we are done. Otherwise, it lies in a disc

of radius

, which meets one of the

having radius

(this is the only reason a disc can be skipped

in our algorithm). Thus the centers lie at a distance

, and so every point at distance less than

from the center of lies at distance at most

from the center of the corresponding

.

A6 Recall the inequalities

(AM-GM) and

(Law of Sines).

Also recall that the area of a triangle with integer

coordinates is half an integer (if its vertices lie at

, the area is

), and that

if and have integer coordinates, then

is an

integer (Pythagoras). Now observe that

and that the first and second expressions are both inte-

gers. We conclude that


26/109

, and so

; that is, is a right angle and

,

as desired.

B1 Notice that

(difference of squares). The latter is easily seen (e.g.,

by AM-GM) to have minimum value 6 (achieved at

).

B2 Consider a triangle as described by the problem; la-

bel its vertices

so that

, lies on

the -axis, and

lies on the line

. Further

let

be the reflection of in the

-axis,

and let

be the reflection of in the line

. Then and

, and so

the perimeter of

is

. It is clear that

this lower bound can be achieved; just set (resp.

)

to be the intersection between the segment and the

-axis (resp. line

); thus the minimum perimeter

is in fact

.

B3 We use the well-known result that the surface area of

the sphere cap

is simply

. (This result is easily verified using

calculus; we omit the derivation here.) Now the desired

surface area is just minus the surface areas of five

identical halves of sphere caps; these caps, up to isome-

try, correspond to

being the distance from the center

of the pentagon to any of its sides, i.e.,

.

Thus the desired area is

(i.e.,

).

B4 For convenience, define

, so that

the given sum is

. If

and are both odd, then

is the sum of an

odd number of s, and thus cannot be zero. Now

consider the case where and have opposite par-

ity. Note that

for all

integers

. Thus

and

; this implies that

is odd, and so

for all . It follows

that

if and

have opposite parity.

Now suppose that

and

are both even.Then

for all , so

can be computed

as twice the sum over only even indices:

Thus

vanishes if and only if

vanishes

(if

, then and have opposite parity

and so

also vanishes).

Piecing our various cases together, we easily deduce

that

if and only if the highest powers of 2

dividing and

are different.

B5 Write

. Then

where

. Now the digits after the deci-

mal point of

are given by

, while the

digits after the decimal point of

are given by

. It follows that the first 1000

digits of are given by

; in particu-

lar, the thousandth digit is .

B6 First solution: Write

. Note

that

and

have the same parity, and recall

that any perfect square is congruent to 0 or 1 (mod 4).

Thus if and

are perfect squares, they are

congruent mod 4. But

(mod4), which is not divisible by 4 if

and

have opposite

parity.

Second solution: We prove more generally that for any

polynomial

with integer coefficients which is not

a perfect square, there exists a positive integer such

that

is not a perfect square. Of course it suffices

to assume

has no repeated factors, which is to say

and its derivative

are relatively prime.

In particular, if we carry out the Euclidean algorithm

on

and

without dividing, we get an in-

teger (the discriminant of

) such that the great-

est common divisor of

and

divides

for any . Now there exist infinitely many primes

such that divides for some : if there were

only finitely many, say,

, then for any di-

visible by

, we have

, that is,

is not divisible

by

, so must be

, but then

takes some

value infinitely many times, contradiction. In particu-

lar, we can choose some such not dividing

, and

choose such that

divides

. Then

(write out the Taylor series

of the left side); in particular, since does not divide

, we can find some such that

is di-

visible by but not by

, and so is not a perfect square.

Third solution: (from David Rusin, David Savitt, and

Richard Stanley independently) Assume that

is a square for all

. For sufficiently large

,

2


27/109

thus if is a large even perfect square, we have

. We conclude this is an

equality of polynomials, but the right-hand side is not a

perfect square for an even non-square, contradiction.

(The reader might try generalizing this approach to ar-

bitrary polynomials. A related argument, due to Greg

Kuperberg: write

as

times a

power series in and take two finite differences to

get an expression which tends to 0 as , contra-

diction.)

Note: in case

has no repeated fac-

tors, it is a square for only finitely many , by a theorem

of Siegel; work of Baker gives an explicit (but large)

bound on such . (I dont know whether the graders

will accept this as a solution, though.)

3


28/109



A-1 Find polynomials

,

, and

, if they exist,such that for all

,

if

if

if

.

A-2 Let

be a polynomial that is nonnegative for all

real

. Prove that for some , there are polynomials

) such that

A-3 Consider the power series expansion

Prove that, for each integer

, there is an integer

such that

A-4 Sum the series

A-5 Prove that there is a constant such that, if

is a

polynomial of degree 1999, then

A-6 The sequence

is defined by

and, for

,

Show that, for all n,

is an integer multiple of .

B-1 Right triangle has right angle at and

; the point is chosen on

so that

; the point is chosen on

so that

.

The perpendicular to

at meets

at

. Evalu-

ate

.

B-2 Let

be a polynomial of degree such that

, where

is a quadratic polynomial and

is the second derivative of

. Show that if

has at least two distinct roots then it must have

distinct roots.

B-3 Let

. For

, let

where the sum ranges over all pairs

of positive

integers satisfying the indicated inequalities. Evaluate

B-4 Let

be a real function with a continuous third deriva-

tive such that

are positive forall

. Suppose that

for all

. Show that

for all

.

B-5 For an integer

, let

. Evaluate the

determinant of the

matrix

, where

is

the

identity matrix and

has entries

for all

.

B-6 Let

be a finite set of integers, each greater than 1.

Suppose that for each integer there is some

such that

or

. Show that

there exist

such that

is prime.


29/109

Solutions to the 60th William Lowell Putnam Mathematical Competition



A1 Note that ifr(x)and s(x)are any two functions, then

max(r, s) = (r+s+ |r s|)/2.

Therefore, ifF(x)is the given function, we have

F(x) = max{3x 3, 0} max{5x, 0} + 3x+ 2= (3x 3 + |3x+ 3|)/2

(5x+ |5x|)/2 + 3x+ 2= |(3x+ 3)/2| |5x/2| x+1

2,

so we may setf(x) = (3x+ 3)/2,g(x) = 5x/2, andh(x) = x+ 12 .

A2 First solution: First factorp(x) = q(x)r(x), whereqhas all real roots andr has all complex roots. Noticethat each root ofq has even multiplicity, otherwise pwould have a sign change at that root. Thusq(x)has asquare roots(x).

Now write r(x) =k

j=1(x aj)(x aj) (possiblebecauser has roots in complex conjugate pairs). Writek

j=1(x aj) = t(x) + iu(x) with t, x having realcoefficients. Then forxreal,

p(x) = q(x)r(x)

=s(x)

2

(t(x) +iu(x))(t(x) +iu(x))= (s(x)t(x))2 + (s(x)u(x))2.

(Alternatively, one can factor r(x) as a product ofquadratic polynomials with real coefficients, write each

as a sum of squares, then multiply together to get a sum

of many squares.)

Second solution: We proceed by induction on the de-

gree ofp, with base case where p has degree 0. As inthe first solution, we may reduce to a smaller degree

in case p has any real roots, so assume it has none.Then p(x) > 0 for all real x, and since p(x) forx

,phas a minimum valuec. Nowp(x)

c

has real roots, so as above, we deduce that p(x) cisa sum of squares. Now add one more square, namely

(

c)2, to getp(x)as a sum of squares.

A3 First solution: Computing the coefficient ofxn+1 in theidentity(12x x2)m=0amxm = 1 yields therecurrencean+1 = 2an + an1; the sequence{an}is then characterized by this recurrence and the initial

conditionsa0 = 1, a1= 2.

Define the sequence {bn} by b2n = a2n1 +

a2n, b2n+1 = an(an1+an+1).Then

2b2n+1+b2n= 2anan+1+ 2an1an+a2n1+a

2n

= 2anan+1+an1an+1+a2n

=a2n+1+a2n = b2n+2,

and similarly 2b2n + b2n1 = b2n+1, so that{bn}satisfies the same recurrence as{an}. Since furtherb0 = 1, b1 = 2 (where we use the recurrence for {an}to calcul

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