CAT 2025 Slot 2 QA Question & Solution
Question
The number of divisors of $(2^{6}\times 3^{5}\times 5^{3}\times 7^{2})$, which are of the form $(3r+1)$, where r is a non-negative integer, is
Options
Solution
The divisors of the given number will have the form $2^a*3^b*5^c*7^d$ with $0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$
Because the divisors should be in the form 3r+1, it cannot be divisible by 3, so (b=0).
Reduce modulo 3: $$2 \text{mod} 3 = 2, 5 \text{mod} 3 = 2, 7 \text{mod} 3 = 1$$
Hence, $$2^a5^c7^d\equiv 2^{a+c}\cdot1^d \equiv 2^{a+c}\pmod3$$
2^k will be in the form 3r+1 only when K is even. So, we need a+c to be even
$$a\in{0,\dots,6}$$ has 4 even, 3 odd values
$$c\in{0,\dots,3}$$ has 2 even, 2 odd
Number of (a,c) with (a+c) even is $$4\cdot2 + 3\cdot2 = 8+6=14$$
For each such pair, there are 3 choices for d = 0,1,2. Thus, total divisors in the form (3r+1) equals $$14\times3=42$$.