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Can I ask someone to take a look at it as I don't understand the solutions provided (or rather I don't understand how they came up with the solutions)? Thanks.

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.

Answer: C.

Hope it helps.

Hi Bunuel,

I noticed that you tried to clarify method 1 below but still not connecting.

How are you drawing the conclusion that it has to be a multiple of 10? Why can't it be 2*175 which is NOT a multiple of 10 or 1*350 etc?

Thanks!

Do we have 175's or 350's to sum? We have 7's and 77's. Try to sum those to get the sum with units digit of 0, and you'll see that the number of terms must be 10, 20, 30, ....
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]

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29 Nov 2014, 14:44

fastest way to solve this IMHO:

350 = 7x50 that mean the sum of term should be 50 if 7 is factored out == 7 (a1+a2+...+an) this can happen if and only if its a factors are multiple of 10 and only C works.

Answer: C

LM wrote:

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?

Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]

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09 Sep 2015, 04:39

I took a while to figure this out on my own but here is how I approached this problem:

Since the answer choices were all over 35, there is a very slight possibility that there will be more than 1 multiple of 77., Therefore I subtracted 77 (multiple of 7) from 350 (multiple of 7) to get 273 (multiple of 7). Divided 273 by 7 to get 39. Getting 40 as the answer. 39*7 + 1*77 = 350.

Even though plugging in the values will not take too much time, however I suggest we try to understand the statement we just derived:

After subtracting the two values, we should have a '0' in the units place ( 50). To get a '0' we need the units place values of both the terms ( 11n and 10x) to be the same.

10x : this term will always have a '0' in the units place irrespective of the value of x --> this means 11n should have a '0' as well and which is possible is 'n' is a multiple of 10.

If each term in the sum \(a_1+a_2+a_3+...+a_n\) 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

Check the solution in attachment

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If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]

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08 Jan 2017, 08:52

I came up with solution using PoE:

let , number of occurrences of 7 is x , number of occurrences of 77 is y , total number of terms is n

7x+77y=350 >> x=50-11y -----(1)

we have , x+y=n

substituting value of x from (1) 50-11y+y=n 50-10y=n As we are subtracting a multiple of 10 from 50 i.e. a number ending with 0 is being subtracted from two digit number ending with 0 Hence , the result should also end with 0

We only have answer choice (C) with such value i.e. 40

Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]

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12 Feb 2017, 08:12

I have found the following approach more direct and less time-comsuming: 7(n-x) + 77x = 350 7n-7x+77x = 350 7(n+10x)=350 n+10x=50 n=50-10x If x=1, then n=50-10=40 (The right option) If x=2, then n=50-20=30 If x=3, then n=50-30=20 If x=4, then n=50-40=10

Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]

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02 Jul 2017, 22:53

I solved it a bit differently. total number of 7's and 77s = 350 Looking at answer choices we can ensure that total number of 7's must be greater than 77's to achieve the range from 38 to 42. Here is next math part: 350 - 77=273. 273/7 = 39 39 (7s)+1 (77) = 40 Hence, Answer choice C.