Hi Roland2rule,
We're asked which of the following CANNOT be a factor of the number 527,P90 where P is the hundred's digit of the number. This question comes down to 'rules of division' - and knowing those rules can save you some time on Test Day.
Mitch's solution is spot-on and ultimately the fastest way to get to the correct answer. Here are the other rules that apply to this question though:
1) A number is evenly divisible by 3 or 9 if the DIGITS of that number sum to a total that is divisible by 3 or 9 (respectively). Here, if P=1, then 527,190 has digits that sum to 24 and since 24 is evenly divisible by 3, then 527,190 is evenly divisible by 3. In that same way, if P=4, then 527,490 has digits that sum to 27 and since 27 is evenly divisible by 9, then 527,490 is evenly divisible by 9.
2) 527,P90 ends in a '0', so it's clearly divisible by 10.
3) There is a 'quirky' divisibility rule for 7s, but it's a bit tougher to apply it here since P takes the place of one of the digits, so brute-force arithmetic might be the easiest way to prove that 7 will divide in. When P=5, 7 evenly divides in to 527,590
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
