CAT 2025 Slot 2 QA Question 12

So, first, the total number of divisors is (6+1)(5+1)(3+1)(2+1) = 7×6×4×3. But I don't need the total; I need only those divisors that are 1 mod 3.

How do I approach this? Maybe I can use the concept of generating functions or modular arithmetic. Since the divisors are formed by multiplying different combinations of the prime factors, I can consider each prime's contribution modulo 3.

Let me break down each prime factor modulo 3:

So, the primes 2 and 5 are congruent to 2 mod 3, 3 is 0 mod 3, and 7 is 1 mod 3.

Now, when forming a divisor, we can ignore the factor of 3 because multiplying by 3^k will only add factors of 3, but since we're looking for numbers ≡1 mod 3, the exponent of 3 in the divisor must be 0. So, we can ignore the 3^5 part because any divisor with 3^k where k>0 will be divisible by 3, hence ≡0 mod 3, which isn't what we want. So, we only consider the exponents of 2, 5, and 7.

So, the problem reduces to finding the number of divisors of 2^6 × 5^3 × 7^2 that are ≡1 mod 3.

Now, let's model each prime's contribution:

So, the total contribution modulo 3 is [(-1)^a × 2^b × 1^c] mod 3. We want this to be 1 mod 3.

So, (-1)^a × 2^b ≡ 1 mod 3.

Let me write this as:

(-1)^a × 2^b ≡ 1 mod 3.

Let me consider the possible values of a and b.

First, note that 2^b mod 3 cycles every 2:

Similarly, (-1)^a alternates between 1 and -1 depending on whether a is even or odd.

So, let's consider cases based on whether a is even or odd.

Then, (-1)^a = 1.

So, 2^b ≡ 1 mod 3.

From above, 2^b ≡ 1 mod 3 when b is even.

So, in this case, b must be even.

Then, (-1)^a = -1 ≡ 2 mod 3.

So, 2^b ≡ 2 mod 3.

Which happens when b is odd.

So, summarizing:

Now, let's find the number of possible (a, b) pairs.

Given that a can be from 0 to 6 (since exponent of 2 is 6), and b can be from 0 to 3 (since exponent of 5 is 3).

First, let's count the number of even a's:

a can be 0,2,4,6 → 4 choices.

For each even a, b must be even: b can be 0,2 → 2 choices.

So, number of pairs when a is even: 4×2=8.

Now, number of odd a's:

a can be 1,3,5 → 3 choices.

For each odd a, b must be odd: b can be 1,3 → 2 choices.

So, number of pairs when a is odd: 3×2=6.

Total number of (a,b) pairs: 8 + 6 = 14.

Now, for each such pair (a,b), c can be anything from 0 to 2, since exponent of 7 is 2.

So, c has 3 choices: 0,1,2.

Therefore, total number of divisors is 14×3=42.

So, the answer is 42, which is option A.

Did I consider all cases correctly?

Yes, I considered a even and odd, and for each case, b accordingly. Then multiplied by the number of c's. Seems correct.

Another way: The total number of divisors without considering mod 3 is (6+1)(3+1)(2+1)=7×4×3=84. But since we're looking for 1 mod 3, which is roughly a third of them, 84/3=28. But wait, that contradicts my earlier answer of 42. Hmm, maybe my initial assumption is wrong.

Wait, no. Because when we fix the exponents of 2,5,7, the 3^0 is fixed, so the total number is (6+1)(3+1)(2+1)=84. But the number of divisors ≡1 mod 3 is not necessarily 84/3 because the distribution might not be uniform.

Wait, but in my earlier calculation, I got 42, which is exactly half of 84. Hmm, that seems conflicting.

Wait, maybe I made a mistake in considering the exponents. Let me think again.

Wait, no, because when we fix the exponents of 2,5,7, each divisor is of