If the multiple of 3 is n and the multiple of 2 is m, the product is 6mn which is to be a product of 10, call it 10p, so 6mn=10p, 3mn=5p. For m, n and p to be integers greater than 1, we can suppose p to be a multiple of 3, for example 3q. Now we have 3mn=15q, so mn=5q. If we suppose m=5, for example, then n=q. The smallest value for n=q=2. We have m=5 and n=2, so the multiple of 3 is 2, making one multiplier 6 and the multiple of 2 is 5, making the other multiplier 10. So the product is 60 which is indeed a multiple of 10. By these conjectures we arrive at (2×3)(5×2)=(6×10). We can interchange m and n, so:
(5×3)(2×2)=(6×10). And we can take this further by introducing a common multiplier on either side of the equation. For example, 2: 2(5×3)(2×2)=2(6×10), then apply this additional multiplier to either m or n:
(10×3)(2×2)=(12×10) or (4×3)(5×2)=(12×10), etc.
Going back to 6mn as the product and m and n are both integers greater than 1, then the minimum product is when m=n=2, that is, 24. If the product is to be a multiple of another number X, 6mn=Xp. If X is not divisible by either 2 or 3, and X is not prime, then p=6, making mn=X. Or p could be any multiple of 6.
If X is prime, then 6mn=Xp so at least one of m or n must be equal to X, so that 6Xn=Xp or 6Xm=Xp, making p=6n or 6m.
If X is divisible by either 2 or 3 (but not both), then we can write X=2Y or X=3Y: 6mn=2Yp or 6mn=3Yp, therefore 3mn=Yp or 2mn=Yp. So p=3 or 2 respectively, and mn=Y.
If X is divisible by 6, then 6mn=6p and mn=p.