adkikani wrote:
Bunuel VeritasKarishma chetan2u gmatbusters niks18 GMATPrepNowI understood above approaches, but almost all above methods requires prime factorization of
three digits numbers and consumed more time. Any alternate approach to solve this efficiently?
Hi
adkikaniN is divisible by 14!. 14! contains only one 11 & 13 and N has to be even because 14! is even
Now look at the option
1690 = 169*10, 169 is square of 13 hence this will contain two 13's. therefore not a factor of N
1210 = 121*10, 121 is square of 11 hence this will contain two 11's. therefore not a factor of N
625 = 625 is square of 25 hence this will contain two 25's. therefore not a factor of N
726, is divisible by 11 (remember the divisibility rule for 11), the resulting number will again be divisible by 11. hence this also contains two or more 11s. Reject this
So your final answer has to be Option E.
If you remember the square and divisibility rule then this question can be solved within 30s by just skimming at the options