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each term equal to 7 or 77

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by DavidG@VeritasPrep » Wed Aug 26, 2015 12:11 pm
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
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by DavidG@VeritasPrep » Wed Aug 26, 2015 12:14 pm
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.
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by rakaisraka » Wed Aug 26, 2015 1:36 pm
Thanks David.
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by [email protected] » Wed Aug 26, 2015 2:09 pm
Hi rakaisraka,

Here is one of the discussions on this particular prompt:

https://www.beatthegmat.com/lucky-again- ... 75428.html

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by Max@Math Revolution » Wed Aug 26, 2015 2:49 pm
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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by Brent@GMATPrepNow » Wed Aug 26, 2015 3:19 pm
rakaisraka wrote: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
The above solutions are faster. However, if you didn't spot them, another approach is to look for a pattern.

The sum of any 1 term will have units digit 7
The sum of any 2 terms will have units digit 4
The sum of any 3 terms will have units digit 1
The sum of any 4 terms will have units digit 8
The sum of any 5 terms will have units digit 5
The sum of any 6 terms will have units digit 2
The sum of any 7 terms will have units digit 9
The sum of any 8 terms will have units digit 6
The sum of any 9 terms will have units digit 3
The sum of any 10 terms will have units digit 0
The sum of any 11 terms will have units digit 7 (at this point, the pattern repeats)

From this, we can conclude that the sum of any 20 terms will have units digit 0
And the sum of any 30 terms will have units digit 0, and so on.

We are told the sum of the terms is 350 (units digit 0), so the number of terms must be 10 or 20 or 30 or . . .

Since C is a multiple of 10, this must be the correct answer.

Cheers,
Brent
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