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

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If each term in the sum a1+a2+a3+.....+an is either 7 or 77 [#permalink] New post 10 Sep 2010, 11:02
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-each-term-in-the-sum-a1-a2-a3-an-is-either-7-or-93974.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Nov 2012, 02:31, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Tough PS problem- please help! [#permalink] New post 10 Sep 2010, 11:15
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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.

Answer: C.

Hope it helps.
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Re: Tough PS problem- please help! [#permalink] New post 11 Sep 2010, 07:05
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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Re: Tough PS problem- please help! [#permalink] New post 11 Sep 2010, 07:16
Same concept alternate solution

Let the number of 77's be X, we know X>=0

7*(n-X) + 77*X = 350
7*(n-X) + 7*X + 70*X = 350
7*n + 70*X = 350

Since X>=0, X can be 0,1,2,3,4,5 .... Correspondingly the value of n would be 50,40,30,20,10,0 (always a multiple of 10)

So the answer in this case is 40.
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Re: Tough PS problem- please help! [#permalink] New post 11 Sep 2010, 20:23
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Let a be the no. of 7s & b be the no. of 77s

7*a+77*b=350
a+11b=50

So, possible values for (a,b) are (39,1);(28,2);(17,3);(6,4)
The no. of terms is (a+b) so, no. of terms can be (40,30,20,10)

After comparing with the given options, 40 is the answer.
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Re: ps question [#permalink] New post 19 Oct 2010, 02:05
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satishreddy wrote:
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.

Hence answer is x+y=40 i.e D.

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in the sum a1 + a2 + a3 + ... + an [#permalink] New post 22 Dec 2010, 21:20
Is there an algebraic way to solve this question rather than picking numbers? 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
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Re: Tough PS problem- please help! [#permalink] New post 23 Dec 2010, 00:43
thanks metallicafan for you supportive help.
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Re: Tough PS problem- please help! [#permalink] New post 28 Dec 2010, 11:25
Hi Bunuel,

Could you provide links to more problems like this?
Thanks!
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Re: Tough PS problem- please help! [#permalink] New post 07 Nov 2012, 15:40
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!
Re: Tough PS problem- please help!   [#permalink] 07 Nov 2012, 15:40
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