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

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VP
Joined: 21 Jul 2006
Posts: 1498
If each term in the sum a1 + a2 + ... + an is either 7 or 77 [#permalink]

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29 Sep 2008, 09:45
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If each term in the sum a1 + a2 + ... + 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

VP
Joined: 05 Jul 2008
Posts: 1379

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29 Sep 2008, 09:56
tarek99 wrote:
If each term in the sum a1 + a2 + ... + 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

I should have looked at the answer before. I started with multiplying 77 with 2. Thought 1 would be too less.

350 can be expressed as 77 * 2 + 28 * 7. looked at answers and I dont see 30.

then lowered the multiple of 77 to 1

77 * 1 = 77 ; 350 -77 = 273, which is divisible by 7 ( 39 times )

Possible value = 40. hence C

On a diff note 77 * 3 + 17 * 7 = 350 ; 77 * 4 + 6 * 7 = 350

Apparently we can't have five 77's in the sequence as the total goes beyond 350.
SVP
Joined: 17 Jun 2008
Posts: 1507

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29 Sep 2008, 12:57
Another approach,

if x is number of 7s in the sequence, then
7x + 77(n-x) = 350

or x + 11(n-x) = 50
or 11n - 10x = 50
or n = (50+10x)/11...now in order for n to be integer, (50+10x) must be a multiple of 110 and possible values of n will be 10,20,30,40,50......

Since, the answer choice contains 40, this will be the correct answer.
VP
Joined: 21 Jul 2006
Posts: 1498

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29 Sep 2008, 15:51
scthakur wrote:
Another approach,

if x is number of 7s in the sequence, then
7x + 77(n-x) = 350

or x + 11(n-x) = 50
or 11n - 10x = 50
or n = (50+10x)/11...now in order for n to be integer, (50+10x) must be a multiple of 110 and possible values of n will be 10,20,30,40,50......

Since, the answer choice contains 40, this will be the correct answer.

Beautiful approach! the OA is indeed C. But I have a question: you said that (50+10x) must be divisible by 110. Did you mean 110 or 11? how did you arrive with 110 and how did you end up with values to be either 10,20,30,40,50, etc? thanks!
Intern
Joined: 29 Sep 2008
Posts: 46

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29 Sep 2008, 16:15
i did the same thing as scthakur, no offence, but why should 10x + 50 be a multiple of 110, ooh.. you mean since it has to have be a multiple of 11 (since you put 11 in denominator) and 10 (since that's a common factor of 10 and 50)

I think its sufficient to say that since 11n = 10(x + 5), n should have 10 as a factor, thus 40 is the only possible option. just trying to make it simple to my small mind.
VP
Joined: 21 Jul 2006
Posts: 1498

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29 Sep 2008, 17:56
aim2010 wrote:
i did the same thing as scthakur, no offence, but why should 10x + 50 be a multiple of 110, ooh.. you mean since it has to have be a multiple of 11 (since you put 11 in denominator) and 10 (since that's a common factor of 10 and 50)

I think its sufficient to say that since 11n = 10(x + 5), n should have 10 as a factor, thus 40 is the only possible option. just trying to make it simple to my small mind.

I agree. I feel that a critical step has been skipped from n = (50+10x)/11. By looking at this equation, I can at least say that (50+10x) is a multiple of 11, but I wasn't sure where to go from there. Where do we go from there exactly?
thanks
VP
Joined: 05 Jul 2008
Posts: 1379

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29 Sep 2008, 18:47
1
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tarek99 wrote:
aim2010 wrote:
i did the same thing as scthakur, no offence, but why should 10x + 50 be a multiple of 110, ooh.. you mean since it has to have be a multiple of 11 (since you put 11 in denominator) and 10 (since that's a common factor of 10 and 50)

I think its sufficient to say that since 11n = 10(x + 5), n should have 10 as a factor, thus 40 is the only possible option. just trying to make it simple to my small mind.

I agree. I feel that a critical step has been skipped from n = (50+10x)/11. By looking at this equation, I can at least say that (50+10x) is a multiple of 11, but I wasn't sure where to go from there. Where do we go from there exactly?
thanks

I can at least say that (50+10x) is a multiple of 11

this means 10 (5+x) is a multiple of 11. Since 11 does not divide 10, 11 must divide (5+x).

so x can be 6, 17, 28, 39 and so on.

At this point I would look down for answers and look for answers in the range of possible values.
VP
Joined: 21 Jul 2006
Posts: 1498

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29 Sep 2008, 19:19
icandy wrote:
tarek99 wrote:
aim2010 wrote:
i did the same thing as scthakur, no offence, but why should 10x + 50 be a multiple of 110, ooh.. you mean since it has to have be a multiple of 11 (since you put 11 in denominator) and 10 (since that's a common factor of 10 and 50)

I think its sufficient to say that since 11n = 10(x + 5), n should have 10 as a factor, thus 40 is the only possible option. just trying to make it simple to my small mind.

I agree. I feel that a critical step has been skipped from n = (50+10x)/11. By looking at this equation, I can at least say that (50+10x) is a multiple of 11, but I wasn't sure where to go from there. Where do we go from there exactly?
thanks

I can at least say that (50+10x) is a multiple of 11

this means 10 (5+x) is a multiple of 11. Since 11 does not divide 10, 11 must divide (5+x).

so x can be 6, 17, 28, 39 and so on.

At this point I would look down for answers and look for answers in the range of possible values.

yeah, you ended that part really nicely. Thanks a lot!
Senior Manager
Joined: 20 Feb 2008
Posts: 292
Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross

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29 Sep 2008, 19:56
From the question stem we get two equations:

7a +77b =350
a+11b=50 ---->1
a+b=n----2

If we try to solve this simultaneously
10b=50-n

Therefore if we substitute the answers in the above equation 50-n must be a multiple of 10 and the only answer that fits this is 40 since (50-40)=10 all other answers when substituted in the the above equation would not give an integer.

So the answer would be C 40
SVP
Joined: 17 Jun 2008
Posts: 1507

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29 Sep 2008, 22:45
tarek99 wrote:
scthakur wrote:
Another approach,

if x is number of 7s in the sequence, then
7x + 77(n-x) = 350

or x + 11(n-x) = 50
or 11n - 10x = 50
or n = (50+10x)/11...now in order for n to be integer, (50+10x) must be a multiple of 110 and possible values of n will be 10,20,30,40,50......

Since, the answer choice contains 40, this will be the correct answer.

Beautiful approach! the OA is indeed C. But I have a question: you said that (50+10x) must be divisible by 110. Did you mean 110 or 11? how did you arrive with 110 and how did you end up with values to be either 10,20,30,40,50, etc? thanks!

I meant 110. The reason is that numerator (50 + 10x) will always be a multiple of 10 no matter what. Hence, numerator can be reduced to 10a for some values of a. Thus, for n to be an integer, a must be a multiple of 11 and hence numerator must be a multiple of 110.

I think I just assumed this step and created confusion.
Intern
Joined: 22 Jul 2009
Posts: 1

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22 Jul 2009, 11:12
1
KUDOS
Please note that the shortest method outwould be to see that addition of 7 and 77 to generate a 0 in the units place 350 would need to hav addition of 7.... ten times i.e in multiples of ten...thus from the given set of options only 40 stands out as such an opbvious option..
ofcourse this method wil nto be applicable if there are more than one option which poses as multiple of 10.
hope this helps.
Director
Joined: 01 Apr 2008
Posts: 848
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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23 Jul 2009, 00:04
jooprajoop wrote:
Please note that the shortest method outwould be to see that addition of 7 and 77 to generate a 0 in the units place 350 would need to hav addition of 7.... ten times i.e in multiples of ten...thus from the given set of options only 40 stands out as such an opbvious option..
ofcourse this method wil nto be applicable if there are more than one option which poses as multiple of 10.
hope this helps.

Short and sweet !!
Re: sequence   [#permalink] 23 Jul 2009, 00:04
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