In how many ways can 144 be written as a product of two DIFFERENT positive whole numbers? $(8\times 18$ and $18\times 8$ count as one way.)

Since $144=2^4\times 3^2$ it has $(4+1)(2+1)=15$ divisors, and these pair off around √144 = 12. Fourteen of them form seven pairs — $1\times 144,2\times 72,3\times 48,4\times 36,6\times 24,8\times 18$ and $9\times 16$ — while 12 is left over, pairing only with itself. The two factors must be DIFFERENT, so $12\times 12$ is barred and the answer is 7. Allowing $12\times 12$ gives the trap 8, which is exactly why the question stresses 'different'; 15 reports the divisor count rather than the number of pairs, and 6 loses the pair $1\times 144$ by treating 1 as not a real factor.