Bunuel
Which of the following could be the sum of 12 consecutive integers?
A 92
B 198
C 328
D 412
E 1,570
Solution:
If the first integer is x, then the second integer is x + 1, the third integer is x + 2, and so on. Therefore, the sum of the 12 consecutive integers, in terms of x, is:
x + (x + 1) + (x + 2) + … + (x + 11) = [x + (x + 11)]/2 * 12 = (2x + 11) * 6 = 12x + 66
Now, let’s check the answer choices starting from choice A:
12x + 66 = 92
12x = 26
x = 26/12 = 13/6 = 2 1/6
Since x has to be an integer, the sum of the 12 consecutive integers could not be 92.
Let’s check choice B:
12x + 66 = 198
12x = 132
x = 132/12 = 11
Since x is an integer, the sum of the 12 consecutive integers could be 198.
Alternate solution:Notice that the sum of the 12 consecutive integers, in terms of x, is 12x + 66, which is a multiple of 6, i.e., a multiple of both 2 and 3. We see that all the answer choices are even; thus, they are all multiples of 2, and therefore, it hinges on which answer choice is also a multiple of 3. Recall that, if the sum of the digits of the number is a multiple of 3, then the number itself is a multiple of 3. Looking at the choices, we see that sums of the digits of the numbers are 11, 18, 13, 7, and 13, respectively, and only 18 is a multiple of 3. Therefore, 198 is the correct answer.
Answer: B