Question Bank - Number of Integral/Positive/Non-Negative Solutions - Set 2
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24 = 2^3 * 3
a * b * c = 24
=> if a = 2^m * 3^n
b = 2^p * 3^q
c = 2^x * 3^y
We can say that m + p + x = 3 and n + q + y = 1
So total solutions = (3 + 3 - 1)C(3 - 1) * (1 + 3 - 1)C(3 - 1)
= 5C2 * 3C2
= 10 * 3
= 30 ways
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We are basically looking for perfect squares between 6 and 101.
9, 16, 25, 36, 49, 64, 81, 100
8 values.
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(x+y)^(y-1) = 100
100 can be represented in p^q form in 2 ways.
case 1: 10^2
(x,y) = (7,3)case 2: 100^1
(x,y) = (98,2)So 2 ways.
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2009 = 41 * 49
√2009 = 7√41
as √41 is irrational it has to be present in both √x and √y.
so we can write, a√41 + b√41 = 7√41
a + b = 7
3 pairs will satisfy - (4, 3), (5, 2) and (6, 1)
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a^2 – b^2 = 288
288 = 2^5 * 3^2
Integer solutions = 2 * Number of factors of (288/4)
288/4 = 72 = 2^3 * 3^2
Number of factors = 4 * 3 = 12
Integer solutions = 2 * 12 = 24
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110 = 2 * 5 * 11
Number of ordered positive triplets = (1 + 3 - 1)C2 * (1 + 3 - 1)C2 * (1 + 3 - 1)C2 = 3 * 3 * 3 = 27
Ordered integer triplets = 4 * 27 = 108 (considering the possibilities (-a, -b, c), (-a, b, -c), (a, -b, -c), (a, b, c))
This includes cases like aab and aaa, which needs to be removed to get the unordered pairs.
a = b = c : no cases possible
a = b or b = c or c = a
As all the prime factors only once, we cannot split it into cases where 2 entries are equal. So the only cases are
(110, 1, 1) & (110, -1, -1) -> which can be arranged in 3!/2! = 3 ways each
All the other triplets can be arranged in 3! ways
So final unordered triplets = (108 - 6)/3! + 6/3 = 102/6 + 2 = 17 + 2 = 19 ways.
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@zabeer answer says 5c2 * 3c2=30 but shouldnt we reduce (1 * 1* 24, 2* 2*6) like in Q7
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@vikas_saini can u plz explain 2nd and 3rd+4th step again. 2nd step:what is -1-what are we reducing in it
3rd step: 3rd+4th step doesnt giv value in 2nd step. what are we missing in it. 3rd+4th= 2T-2
