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317 - · PDF fileIf a, b, c are + ve real numbers, then show that a2b + ab2 +c2a + ca2 + b2c +...
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Olympiad (RMO)-2011 Sample paper-1 Solutions “Difference between Dreams and Goals is One’s WILL Power” M. Marks: 80 M. Time: 8 8 = 64 min 1. Find the number of ways in which 39312 can be resolved into two factors which are prime to each other. Sol: 39312 = 2 4 × 3 3 × 7 × 13 There are four distinct primes in factorization of the given number. The required number of ways 2 3 = 8 2. Show that 1 1997 + 2 1997 +….+1996 1997 is divisible by 1997. Sol: We shall make groups of the terms (1 1997 +1996 1997 ) + (+ 2 1997 + 1995 1997 ) +…. + (1998 1997 + 999 1997 ) Here each bracket is of the form 2n 1 2n 1 i i a b . It divisible by (ai + bi) But ai + bi 1997 for all i Each bracket and hence their sum is divisible by 1997. 3. Find the last two digit of 3 1997 . Sol: This is same as asking what is remainder when 3 1997 100 3 4 81 mod 100 3 8 61 mod 100 3 12 41 mod 100 3 16 21 mod 100 3 20 1 mod 100 Now, 3 40 , 3 60 , 3 80 3 100 , …., 3 1980 all are 1 mod 100 We know 3 16 21 mod (100) 3 17 21 × 3 mod 100
Transcript of 317 - · PDF fileIf a, b, c are + ve real numbers, then show that a2b + ab2 +c2a + ca2 + b2c +...
Olympiad (RMO)-2011
Sample paper-1
Solutions
“Difference between Dreams and Goals is One’s WILL Power”
M. Marks: 80 M. Time: 8 8 = 64 min
1. Find the number of ways in which 39312 can be resolved into two factors
which are prime to each other.
Sol:
39312 = 24 × 33 × 7 × 13
There are four distinct primes in factorization of the given number.
The required number of ways 23 = 8
2. Show that 11997 + 21997 +….+19961997 is divisible by 1997.
Sol:
We shall make groups of the terms (11997 +19961997) + (+ 21997 + 19951997) +…. +
(19981997 + 9991997)
Here each bracket is of the form 2n 1 2n 1i ia b . It divisible by (ai + bi)
But ai + bi 1997 for all i
Each bracket and hence their sum is divisible by 1997.
3. Find the last two digit of 31997.
Sol:
This is same as asking what is remainder when 31997 100
34 81 mod 100
38 61 mod 100
312 41 mod 100
316 21 mod 100
320 1 mod 100
Now, 340, 360 , 380 3100, …., 31980 all are 1 mod 100
We know 316 21 mod (100)
317 21 × 3 mod 100
317 63 mod 100
31997 = 31980 × 317
31980 1 mod 100
and 317 63 mod 100
31997 63 mod 100
Last two digit is 63.
4. If a, b, c are + ve real numbers, then show that a2b + ab2 +c2a + ca2 + b2c + bc2 + 2abc
8abc
Sol:
Let S = a2b + ab2 +c2a + ca2 + b2c + bc2 + 2abc
Factorizing we get S =(b + c) (c + a) (a + b)
We have a + b 2 ab
b c 2 bc and a c 2 ac
Multiplying these we get
(a + b) (b + c) (a + c) 8abc
S 8abc
Hence proved.
5. If a, b, c are + ve real number, prove that 6abc a2 (b + c) + b2 (c + a)+ c2
(a + b) 2(a3 + b3 + c3)
Sol:
we have a2 + b2 > 2ab
a2 + b2 – ab > ab
(a + b) (a2 + b2 –ab) > ab (a + b)
a3 + b3 > ab(a + b)
similarly, b3 + c3 > bc (b + c) c3 + a3 > ca(c + a)
Adding all these we get
2(a3 + b3 + c3) > ab (a + b) + bc(b + c) + ca(c + a)
Again A.M. > G.M.
ab a b bc b c ca c a
6
1
2 2 2 2 2 2 6a b ab b c bc c a ca
ab(a + b) + bc (b + c) + ca(c + a) > 6abc
Hence
6abc a2 (b + c)+ b2(c + a) + c2 (a + b) 2(a3 + b3 + c3)
6. In how many ways can the letters of the word JUPITER be arranged in a row so that
the vowels will appear in alphabetic order?
Sol:
JUPITER has 7 letters having 3 vowels U, I, E and 4 consonants J, P, T, R.
Alphabetic order of vowels is E, I, U.
The condition alphabetic order to vowels implies that E can appear in 1st, 2nd, 3rd
or 5th place.
In every arrangement of 3 vowels, the 4 consonants can be placed in the remaining 4
places in
4 3 2 1 = 24 ways
We consider following 5 cases.
Case I. If E appears in 1st place, then I and U will appear in following 15 ways
(2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (3, 4), (3, 5), (3, 6),
(3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7) and (6, 7)
Total number of arrangement is
15 24 = 360
Case II. If E appears in 2nd place, then I and U Will appear in following 10 ways.
(3, 4), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7).
Total number of arrangement is
10 24 = 240
Case III. If E appears in 3rd place, then I and U will appear in following 6 ways.
(4, 5), (4, 6), (4, 7), (5, 6), (5, 7), (6, 7) .
Total number of arrangement is
6 24 = 144
Case IV. If E appears in 4th place, then I and U will appear in following 3 ways.
(5, 6), (5, 7) and (6, 7)
Total number of arrangement is
3 24 = 72
Case V. If E appears in 5th place, then I and U will appear in 1 way
(6, 7)
Total number of arrangement is
1 24 = 24
By Addition rule, total number of arrangement is
360 + 240 + 144 + 72 + 24 = 840.
7. How many +ve integers n are there such that n is a divisor of one of the numbers 1040
, 2030 ?
Sol.
(a1 + 1) (a2 + 1) ……..(ak + 1)
If a1 a2 ak1 2 kn p p .........p
where p1……..pk are integers.
Now, a =1040 = 240 540
b = 2030 = 260 530
gcd of a, b is c = 240 530
Let A, B denote the sets of divisor of a, b respectively.
Then A B is set of divisors of c. 2A 41
B 61 31
A B 41 31
Hence A B 1681 1891 1271 2301
8. A student on vacation for d days observed that
(i) it rained 7 times morning or afternoon
(ii) when it rained in the afternoon it was clear in the morning
(iii) there were 5 clear afternoons, and
(iv) there were 6 clear mornings. Find d.
Sol.
Let the set of days it rained in the morning be M2
Let Ar be the set of days it rained in afternoon.
M’ be the set of days, when there were clear morning.
Ar’ be the set of days when there were clear afternoon.
By condition (ii), wer get r rM ' A
By (iv), we get Mr’ = 6
By (iii), we get Ar’ = 5
By (i), we get r rM A 7
Mr and Ar are disjoint sets
n(Mr) = d 6
n (Ar) = d 5
Applying the principle of inclusion-exclusion
r r r r r rn(M A ) n(M ) n(A ) n(M A )
7 (d 6) (d 5) 0
2d 18
d 9
