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If each term in the sum a1+a2+a3+.....+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

Number of approaches are possible.

For example: as units digit of 350 is zero then # of terms must be multiple of 10. Only answer choice which is multiple of 10 is C (40).

To illustrate consider adding:

*7 *7 ... 77 77 ---- =350

So, several 7's and several 77's, note that the # of rows equals to the # of terms. Now, to get 0 for the units digit of the sum the # of rows (# of terms) must be multiple of 10. Only answer choice which is multiple of 10 is C (40).

Answer: C.

Or: \(7x+77y=350\), where \(x\) is # of 7's and \(y\) is # of 77's, so # of terms \(n\) equals to \(x+y\);

\(7(x+11y)=350\) --> \(x+11y=50\) --> now, if \(x=39\) and \(y=1\) then \(n=x+y=40\) and we have this number in answer choices.

Nice explanation. I was clueless after reading question and was not able to decide on how to create the equation. I like first deduction which needs some presence of mind _________________

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in each in the sum a1+a2+..................+an is either 7 or 77 and the sum equals to 350, which of the following could be equal to n

38 39 40 41 42

Let the number of terms having value 7 is x ,and the number of terms having value 77 is y.

Now according to question,

7x+77y=350 or, 7x + 11*7 y =350 or, 7(x+11y) =350 or x+11y = 50

If y =1 and x = 39 , x+11y=50 (this is the only case we can take otherwise in all other case (x+y) will be less than or equal to 29 and their is no option in answer which is having less than 29 value.

You know that there have to be a lot of 7ns in the answer, because the answer choices are significant. You also know that 350 = 50*7 however this is not an answer choice. But this is the key to the answer 350 - 11*7 or 77 = 39*7 so the answer has to be 39*7 and 1*77 40!

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