There are 27 different threedigit 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
OA E
Source: GMAT Prep
There are 27 different threedigit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the
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This is a great example of how the GMAT often rewards students for thinking outside the box.BTGmoderatorDC wrote: ↑Thu Apr 15, 2021 6:10 pmThere are 27 different threedigit 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
OA E
Source: GMAT Prep
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 3digit 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.