CAT 2025 Slot 2 · Quant · Question 11

A divisor of the form 3r+1 is one that leaves remainder 1 on division by 3.

Any divisor containing a factor 3 leaves remainder 0, so the exponent of 3 must be 0.

A general divisor is therefore {2}^{p}{5}^{q}{7}^{s} with 0\le p\le 6, 0\le q\le 3 and 0\le s\le 2.

Modulo 3, 2\equiv -1, 5\equiv -1 and 7\equiv 1.

So the divisor is \equiv{\left( -1 \right)}^{p}{\left( -1 \right)}^{q}{\left( 1 \right)}^{s}={\left( -1 \right)}^{p+q}, which equals 1 exactly when p+q is even.

p has 4 even choices and 3 odd; q has 2 even and 2 odd.

Both even gives 4\times 2=8 pairs and both odd gives 3\times 2=6, so 14 in all.

The power of 7 never affects the remainder, so all 3 choices are free: 14\times 3=42.

Shortcut. Since 7 is 1 mod 3 it is irrelevant, and 2 and 5 are both -1 mod 3, so they behave identically. You are simply counting \left( p,q \right) pairs with the same parity out of 7\times 4=28 — which is exactly half, 14 — then multiplying by 3.