If each term in the sum a1 + a2 + a3...+ an is either 7 or 77 and the sum equals 350, which of the following could be n?

A) 38
B) 39
C) 40
D) 41
E) 42


The way I solved this is:

7x + 77y = 350 (where x = number of 7's and y = number of 77s)
7(x + 11y) = 350
x + 11y = 50

You also know that x + y = n.

If you subtract the two equations you will get: 10y = 50 - n
Since y MUST be an integer, the only possible options of n = 10, 20, 30, 40, etc....

Is there a better way of solving this question? I did not think of this during the exam...but am thinking there may be an easier way with better logic.

Any input?

Thanks!

I was running out of time when I got this question on the test, I am not sure if this is the best way to solve it, still quicker than conventional approach..

Multiplied all of the given answer choices by 7

A) 38*7= 266
B) 39*7= 273
C) 40*7= 280
D) 41*7= 287
E) 42*7= 294

I tried to calculate the no. of times 7 has occured in the sequence. If you watch carefully, only B will lead to the sum of 350 if 77 is added to it.

So total no. of occurences of 7's - 39 and 1 occurence of 77 is possible in the available answer choices.

So n = 40.

GMAT 2007

I got it on my test too and here is what I did.

Since, there is no 50 in the answer choices (350/7 = 50), we know there is atleast one 77.

350 - 77 = 273
273/7 = 39
39+1 = 40. Bingo!

If 40 wasn't there, I would have subtracted 77 from 273 and continued in a similar way.

Good approach givemeanid. Way to go!!

Thanks for the help guys. Both approaches seem more efficient than mine.