Number properties problem - GMAT Prep Practice test

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Each term in the sum a1+a2+...+an is either 7 or 77, and the sum equals 350. Which of the following could be equal to n?
A. 38
B. 39
C. 40
D. 41
E. 42

Quantitative question from GMAT Prep Practice Test.

Can somepne please explains how you approach the answer with this question? This kind of number properties question sometimes makes me blank, I don't really know where to start.. :ss
Thanks a lot!
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by Jim@StratusPrep » Thu Dec 06, 2012 8:16 am
Start with all 7s. There would have to be 50 of them. In order to add one 77 at this point you would have to get rid of eleven 7s. Essentially you would eliminate 10 numbers so your answer is 40
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by Brent@GMATPrepNow » Fri Dec 07, 2012 3:07 pm
dtl wrote:Each term in the sum a1+a2+...+an is either 7 or 77, and the sum equals 350. Which of the following could be equal to n?
A. 38
B. 39
C. 40
D. 41
E. 42
Looking for a pattern is another possible route:

Since both 7 and 77 have 7 as their units digit, we know that if we take any two terms, their sum will have a units digit of 4 (e.g., 7 + 7 = 14, 7 + 77 = 84, 77 + 77 = 154)

Similarly, if we take any three terms, their sum will have a units digit of 1. (e.g., 7 + 7 + 7 = 21, 7 + 7 + 77 = 91, etc.)

Now let's 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

Another option is to try some different configurations. If we do, we see that adding 39 7's and 1 77 gives us a total of 350. (39 + 1 = [spoiler]40 = E[/spoiler])
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by ritind » Sat Dec 08, 2012 3:07 am
Looking at the answers choices you can make out that no of 7s are more than no of 77s
assume if there was only one 77 then 350-77 = 273
divide 273 by 7 = 39
no of 7's - 39
no of 77 - 1
total terms - 40