Problem solving in algebra and numbers

Compiled by Gary Tsang

January 31, 2010

Problem 1. Mary and David play the following number-guessing game. Marywrites down a list of positive integers x1,x2, . . . ,xn. She does not reveal themto David but only tell him the value n. David chooses a list of positive integersa1,a2, . . . ,xn and ask Mary to tell him the value of a1x1 +a2x2 + · · ·+anxn. ThenDavid chooses another list of positive integers b1,b2, . . . ,bn and asks Mary for thevalue of b1x1+b2x2+ · · ·+bnxn. Play continues in this way until David is able todetermine Mary’s numbers. Find the least number of rounds that David needs inorder to determine Mary’s number.

Solution : Answer: 2 rounds. First of all, David choose a1 = a2 = · · · = an = 1so that he obtains the value m = x1 + x2 + · · ·+ xn, then he choose b1 = 1,b2 =m+1,b3 = (m+1)2, . . . ,bn = (m+1)n−1 and obtain the value m′ = b1x1+b2x2+· · ·+bnxn = x1 +(m+1)x2 +(m+1)2x3 + · · ·+(m+1)n−1xn which is the base-(m+ 1) representation of m′. By the uniqueness of base-n representation of aninteger, David can determine x1,x2, . . . ,xn.Alternative: Once Davaid obtains the value m = x1+x2+ · · ·+xn and let 10k−1 ≤m < 10k, he can choose b1 = 1,b2 = 10k,b3 = 102k, . . .bn = 10(n−1)k and he canstill determine the value x1,x2, . . . ,xn.

Problem 2. Suppose P(x) is a polynomial with integer coefficients and P(a) =P(b) = P(c) = P(d) = 2 for distinct integers a,b,c and d. Which of the followingis true?a. There is no integer k such that P(k) = 1,3,5,7 or 9.b. There is an integer k such that P(k) = 1,3,5,7 or 9.c. Neither a. nor b. is true in general and there are examples for both.

Solution : Answer: a. Proof: Since P(a) = P(b) = P(c) = P(d) = 2, then we canlet P(x) = (x−a)(x−b)(x−c)(x−d)Q(x)+2.Since Q(x) is an integer and a,b,cand d are distinct, therefore P(x)−2 must have at least 4 distinct integral factors:(x−a), (x−b), (x−c) and (x−d). Which is impossible if P(k) = 1,3,5,7,9.

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Problem 3. Find all integers solutions to the system of equations:

x3 + y3 + z3 = x+ y+ z = 3

Solution : (From Kin Y. Li, Math Problem) Consider

(x+ y+ z)3− x3− y3− z3 = 3(x+ y)(y+ z)(z+ x)8 = (x+ y)(y+ z)(z+ x)

23 = (3− x)(3− y)(3− z)

By checking the factorization of 8, and note that (3− x)+ (3− y)+ (3− z) = 6,we have {x,y,z}= {1,1,1} and {4,4,−5}.

Problem 4. Suppose P(x) is a polynomial of degree 8 with real coefficients andP(k) = 1

k for k = 1,2,3, . . . ,9. Determine the number P(10).

Solution : Since kP(k)− 1 = 0 for k = 1,2,3, . . . ,9. Thus xP(x)− 1 = a0(x−1)(x−2) · · ·(x−9) for some constant a0.Because P(x) is a polynomial, we needthe constant term of a0(x−1)(x−2) · · ·(x−9)+1 equal 0. Thus a0 =

19! and so

P(x) = (x−1)(x−2)···(x−9)+9!9!x and so P(10) = 1

2 .

Problem 5. Find the smallest integer B such that B > (√

2+√

3)6.

Solution : Note that (√

3+√

2)(√

3−√

2) = 1 and 0 < (√

3−√

2)n < 1 for alln∈N. Now we let sn = (

√3+√

2)2n+(√

3−√

2)2n = (5+2√

6)n+(5−2√

6)n,One can show that sn satisfies the recurrent relation sn = 10sn−1− sn−2. By directcalculation we can get s1 = 10 and s2 = 98. Thus s3 = 10(98)− 10 = 970. Thisyields (

√3+√

2)6 +(√

3−√

2)6 = 970 and B = d(√

3+√

2)6e= 970.

Problem 6. Problem: Given that a, b and c are positive integers and

abc+ab+bc+ ca+a+b+ c = 2003

Find the least possible value of abc.

Solution : Note that

abc+ab+bc+ ca+a+b+ c = 2003(a+1)(b+1)(c+1) = 2004

Since a, b and c are positive integers, we factorize 2004 and only consider thefactors greater than 1:

(a+1)(b+1)(c+1) = 22×3×167

To minimize abc, we need as many 1 as possible, so choose (a+1)(b+1)(c+1)=2×2×501, then {a,b,c}= {1,1,500} and so minimum value of abc = 500.

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