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?
Sequence Problem
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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.
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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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.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?
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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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 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.
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...yes!!! That's also the fastest way to solve this problem: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?
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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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.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.
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