CAT 2025 Slot 2 · Quant · Question 20

625={5}^{4} and 128={2}^{7}.

Applying the exponents, {\left( 625 \right)}^{65}={5}^{4\times 65}={5}^{260} and {\left( 128 \right)}^{36}={2}^{7\times 36}={2}^{252}.

The product is N={5}^{260}\times{2}^{252}.

Pairing each 2 with a 5 makes a 10, and there are 252 twos, so N={5}^{8}\times{10}^{252}.

Since {5}^{8}=390625, the number is 390625 followed by 252 zeros.

Zeros contribute nothing, so the digit sum is 3+9+0+6+2+5=25.

Shortcut. Whenever a product is a power of 2 times a power of 5, strip off {10}^{\min} immediately — the trailing zeros can never affect a digit sum. All that matters is the leftover {5}^{8}.