GMAT Prep URGENT!!!

[Stendulkar]

If 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 :

1) 38
2) 39
3) 40
4) 41
5) 42

[Brent@GMATPrepNow]

If 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 :

1) 38
2) 39
3) 40
4) 41
5) 42

I recently answered this same question. My post is below.

Looking for a pattern is one 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.


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 = 40)

[Stendulkar]

Thanks a lot Brent....this was a very easy question...I made a silly mistake...

I also followed the same method. I subtracted 77 from 350 which should have yielded 273 ( i.e. 7 * 39). However I got 263 in a hurry which is not divisible by 7. Then i subtracted 154 ( 2 *77)from 350, got 196 ( 28*7) and concluded that n should be 30...not in the options...

Thank You for your time. I will be posting some more problems. Will appreciate if you can address them.

Just to get an understanding of the test...at wht level would you rate this question?

[Brent@GMATPrepNow]

Just to get an understanding of the test...at wht level would you rate this question?

I'd say this is a mid-range question, mainly because someone could also solve it by testing different combinations of numbers.

[Stendulkar]

Thank You Brent...I have posted some questions in the DI section...please go through them..

[deepakh88]

I had answered this question on another forum. Here goes my solution.
350/7 =50.
Implies, it takes 50 7s and 0 77s to get a 350. Possible n = 50.
Considering 77 is 11 7s put together, try 1 77 and 39 7s (50-11 7s).
There u go. 1+39 = 40.

[vkb001]

I did it this way:

The sequence has a number of 7's and 77's. So, the sum of the sequence is something like this:

7x + 77(n-x) = 350, where x is the number of 7's in the sequence.

=> 77n + 7x - 77x = 350
=> 11n - 10x = 50
=> x = (11n - 50)/10

Now, we know that x is an integer. Therefore, it should be divisible by 10. This is possible only when last digit of n is 0. From the given list of options, 40 is the answer.