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Sequence Problem

by vongdn » Wed Oct 06, 2010 7:59 pm
If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum = 350, which of the following could be equal to n?

38
39
40
41
42


Answer is 40

Is this because the only way that the sums of 7 and 77 can come up to 350 is if they add up to a sum that ends in 0, and that the only multiple of 7 that ends in 0 is if it were multiplied by 10?
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by Rahul@gurome » Wed Oct 06, 2010 8:12 pm
If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum = 350, which of the following could be equal to n?

Let the number of terms which are 77 be x.
So number of terms which are 7 is (n-x).
Or 77x + 7(n-x) = 350.
Or 70x + 7n = 350.
Or n = 50 -10x.
Note that n and x are non negative integers. Also n is more than 0. So 50 -10x > 0 or x < 5.
Now if x is 1, n = 40.

So n can take the value of 40.
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by GMATGuruNY » Wed Oct 06, 2010 8:16 pm
vongdn wrote:If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum = 350, which of the following could be equal to n?

38
39
40
41
42


Answer is 40

Is this because the only way that the sums of 7 and 77 can come up to 350 is if they add up to a sum that ends in 0, and that the only multiple of 7 that ends in 0 is if it were multiplied by 10?
350/7 = 50. So if each term were 7, we'd have 50 terms. The answer choices are all a little less than 50. So most -- but not all -- of the terms will be 7.

If 1 term = 77, we have 350-77 = 273 left.
273/7 = 39. This works!
So 1 term = 77, 39 terms = 7.
Total number of terms is 1+39 = 40.

The correct answer is C.
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by saurabhmahajan » Wed Oct 06, 2010 11:06 pm
GMATGuruNY,

I have seen many of your posts where you have solved by plugging numbers and inspired form that i have also solved many of the problems on this forum the same way as you did.

Thanks for helping me and inspiring me.
Keep posting.
Thanks and regards,
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by Testluv » Thu Oct 07, 2010 2:31 am
Is this because the only way that the sums of 7 and 77 can come up to 350 is if they add up to a sum that ends in 0, and that the only multiple of 7 that ends in 0 is if it were multiplied by 10?
...yes!!! That's also the fastest way to solve this problem:

There's eleven 7s in 77. So how can a bunch of 7s sum to 350? Well, since 350 ends in "0", the number of 7s must be a multiple of 10. Only one choice is a multiple of 10--choice C!
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by GMATGuruNY » Fri Oct 08, 2010 10:44 am
saurabhmahajan wrote:GMATGuruNY,

I have seen many of your posts where you have solved by plugging numbers and inspired form that i have also solved many of the problems on this forum the same way as you did.

Thanks for helping me and inspiring me.
Keep posting.
I'm glad that you've been finding my posts helpful. It's my goal -- and that of The Princeton Review -- to use and teach consistent methods that can be applied to many kinds of problems. I plug in my own values or plug in the answer choices in order to solve a majority of GMAT questions.
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by Yanat » Fri Oct 08, 2010 10:20 pm
350-77 = 273

273/7 = 39

so we have 39+1 = 40 and so it is C.
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