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:

• A. \(42\)
• B. \(36\)
• C. \(56\)
• D. \(24\)

Hint:

When counting divisors with a condition like “of the form $3r+1$”, work modulo 3: \begin{itemize} \item Express each prime factor modulo 3. \item Eliminate exponents that force the divisor to be 0 mod 3. \item Use parity or congruence conditions on exponents to match the required remainder. \end{itemize}

Answer 1

The Correct Option is A

Approach: Instead of parity-counting, build the answer multiplicatively. Group the divisors as (part from $2^a5^c$) $\times$ (part from $3^b7^d$) and track each piece's residue mod 3.

Step 1: A divisor is $2^a3^b5^c7^d$. Mod 3 the primes reduce to $2\equiv 2,\ 3\equiv 0,\ 5\equiv 2,\ 7\equiv 1$. Any factor of 3 kills the residue, so $b=0$ is compulsory and $7^d\equiv 1$ always.

Step 2: So the residue depends only on $2^a5^c \equiv 2^{a+c}\pmod 3$. The powers of 2 cycle $2^0\equiv1,\ 2^1\equiv2,\ 2^2\equiv1,\dots$ i.e. $2^{a+c}\equiv 1$ when $a+c$ is even, $\equiv 2$ when odd.

Step 3 (clean counting trick): Total divisors with $b=0$ are $(7\text{ values of }a)\times(4\text{ values of }c)=28$, ignoring $d$. Among these I need $a+c$ even. List by $a$-parity: $a$ even (4 ways) needs $c$ even (2 ways) $=8$; $a$ odd (3 ways) needs $c$ odd (2 ways) $=6$. So $14$ of the $28$ qualify.

Step 4: Re-attach the free $7^d$ with $d\in\{0,1,2\}$, each keeping the residue $1$: multiply by $3$. \[ 14\times 3 = 42. \]

Answer: 42.