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by magical cook » Mon Feb 04, 2008 1:22 pm
If the number 52,1n9, where n represents the tens digit, is a multiple of 3, then the value
of n could be which of the following?
A. 6
B. 5
C. 3
D. 1
E. 0
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by smar83 » Mon Feb 04, 2008 2:35 pm
is the answer "D" ?
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by Stuart@KaplanGMAT » Mon Feb 04, 2008 2:39 pm
Quick trick for deciding if a number is a multiple of 3:

if the sum of the digits of the number is a multiple of 3, then the whole number is a multiple of 3.

Here we have:

52,1n9 and know that the number is a multipe of 3.

Therefore, 5+2+1+n+9 = a multiple of 3

17 + n = a multiple of 3.

Looking at the choices, only

(d) 1

will give us what we want (since 17+1=18, which is a multipe of 3).
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by luvaduva » Sat Feb 09, 2008 3:33 pm
Here is a great link for divisibility rules:

https://www.mathwarehouse.com/arithmetic ... -tests.php

It has 2,3,4,5,6,7,8,10, and 11
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by bhumika.k.shah » Fri Jan 22, 2010 11:57 am
Again Backsolving helps :-)
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by Concordio » Sat Jan 23, 2010 8:30 pm
I also used backsolving, but knowing the divisibility rule would have been much faster. I would've used the divisibility rule if I didn't doubt myself due to 9 having the same divisibility rule.
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