How many three digit numbers exist such that the third digit is the geo mean 
How many three digit numbers exist such that one of the digits is the geometric mean of the other two?
  A 12, B 18, C 24, D other

So, $N = 100a + 10b + c$ let $c =\sqrt{ab}$. 
$ab$ must be a perfect square, so $ab = 1, 4, 9, 16, 25, 36, 49, 64, 81$
Case: $ab = 1 \implies a=1, b=1$. 
Case: $ab = 4 \implies a = 2, b=2$
Case: $ab = 9 \implies a=3, b=3$.
Case: $ab = 16 \implies (a=8, b=2), (a=4, b=4)$
Case: $ab = 25 \implies a=5, b=5$
Case: $ab = 36 \implies (a=9, b=4), (a=6, b=6)$
Case: $ab = 49 \implies (a=7, b=7)$
Case: $ab = 64 \implies (a=8, b=8)$
Case: $ab=81 \implies (a=9, b=9)$.
$(a=8, b=2, c = 4)$ also means that we could replace this with, $ (a, b, c) = (2, 4, 8) = (4, 2, 8) = (4, 8, 2)$
$(a=9, b=4, c=6)$ also means $(a, b, c) = (9, 4, 6) = (4, 6, 9) = (9, 6, 4) = (4, 9, 6) = (6, 9, 4) = (6, 4, 9)$.
So to me, the answer is B $18$

Third digit is $\sqrt{ab}$ implies square root should yield single digit so it can be $[0,9]$ thus now digits whose product is 0 so $100,..900$ , for 1 is only $1,1$ so number $111$ then for 2 so product is $4$ thus numbers are $222,142,412$ then $3$ so $333,193,913$ now $4$ so $284,824,444$ then $5$ so $555$ then $6$ so $496,946,666,$  for $7$ so $777$ then $8$ so $888$ and finally $9$ so $999$ thus total numbers are $9+1+3+3+3+1+3+1+1+1=26$ so its $D$
