Question Bank  Number of Integral/Positive/NonNegative Solutions  Set 2

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

We are basically looking for perfect squares between 6 and 101.
9, 16, 25, 36, 49, 64, 81, 100
8 values.

(x+y)^(y1) = 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.

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)

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

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

@vikas_saini can u plz explain 2nd and 3rd+4th step again. 2nd step:what is 1what are we reducing in it
3rd step: 3rd+4th step doesnt giv value in 2nd step. what are we missing in it. 3rd+4th= 2T2