Carcass wrote:
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
A. 2,704
B. 2,990
C. 5,404
D. 5,444
E. 5,994
This is a great example of how the GRE often rewards students for thinking outside the box.
Here, we can apply a divisibility rule that says integer N is divisible by 3 if and only if the sum of the digits of N is divisible by 3.
For example, we know that 11112 is divisible by 3, because 1+1+1+1+5=9, and 9 is divisible by 3.
Notice that 1+2+3 = 6, and 6 is divisible by 3
This means any 3-digit integer consisting of a 1, a 2 and a 3 must be divisible by 3
If each of the 27 integers in the sum is divisible by 3, then the sum must also be divisible by 3.
In other words, the correct answer must be divisible by 3, which means the sum of its digits must be divisible by 3.
Let's check each answer choice...
A. 2+7+0+4 = 13, which is not divisible by 3.
Eliminate. B. 2+9+9+0 = 20, which is not divisible by 3.
Eliminate. C. 5+4+0+4 = 13, which is not divisible by 3.
Eliminate. D. 5+4+4+4 = 17, which is not divisible by 3.
Eliminate. E. 5+9+9+4 = 27, which IS divisible by 3.
By the process of elimination, the correct answer is E.
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Brent Hanneson - founder of Greenlight Test Prep