CAT 2023 Slot 2 · Quant · Question 3

If a number divides two quantities, it also divides every whole-number combination of them.

So k divides any expression of the form \alpha \left( m+2n \right)+\beta \left( 3m+4n \right) with integer \alpha and \beta.

Choose the combination that wipes out m: take three times the first minus the second.

That gives 3\left( m+2n \right)-\left( 3m+4n \right)=3m+6n-3m-4n=2n.

So k divides 2n.

Now choose the combination that wipes out n: take twice the first minus the second.

That gives 2\left( m+2n \right)-\left( 3m+4n \right)=2m+4n-3m-4n=-m.

So k divides m.

Hence k is guaranteed to be a common divisor of m and 2n — the second option.

A concrete check: m=2, n=1 gives m+2n=4 and 3m+4n=10, so k=2 works; here k divides m=2 and 2n=2, but not n=1, which rules out the fourth option.

Shortcut. Eliminating one variable at a time from the two given expressions is the whole method, and the coefficients needed are small. The example m=2, n=1, k=2 alone kills options 3 and 4, and option 1 fails for m=1, n=2, k=5.