Each term in the sequence A1 +A2 + A3 ----- + An is either 7 or 77 and th sum equals 350. What could be the value of n.
a) 38
b) 39
c) 40
d) 41
e) 42

Let's take the simplest possible scenario. Imagine that every term in the sequence were 7. How many of those would we need for the sum to be 350? We'd need 50. Not an answer choice, but it is in the ballpark.

Say, then that we had one 77. We'd need the 7's to sum to 273. (350 - 77 = 273) There are thirty-nine 7's in 273. If we have one 77 and and thirty-nine 7's, we'll have a total of 1 + 39 = 40 terms. Answer is C

Each term in the sequence A1 +A2 + A3 ----- + An is either 7 or 77 and th sum equals 350. What could be the value of n.
a) 38
b) 39
c) 40
d) 41
e) 42

One alternative: use the units digits. Both 77 and 7 end in 7. We know that the sum should be 350, which ends in 0. Now use the answer choices.

A) If we had 38 terms all ending in 7, the units digit would be 6. (8*7 = 56, and 56 ends in 6.) No good. 350 ends in 0.

B) If we had 39 terms all ending in 7, the units digit would be 3. (9*7 = 63, and 63 ends in 3.) No good

C) If we had 40 terms all ending in 7, the units digit would be 0. (0*7 = 0) 350 ends in 0, so we've got our answer.

Thanks David.

Each term in the sequence A1 +A2 + A3 ----- + An is either 7 or 77 and th sum equals 350. What could be the value of n.
a) 38
b) 39
c) 40
d) 41
e) 42


Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


==> 350=7*50=7+7+....+7 (the number of 7s is 50), 77=7*11=7+7+...+7 (the number of 7s is 11)

so,
350=7*50 ==> n=50

350=7*39+7*11=7*39+77*1 ==> n=40=39+1

350=7*28+7*22=7*28+77*2 ==> n=30=28+2

50, 40, 30 coule be the value of n

Therefore, the answer is C.

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