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

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Joined: 20 Sep 2005
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If each term in the sum a1+a2+a3+.....+an is either 7 or 77 [#permalink]

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22 Mar 2006, 23:07
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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 30 Oct 2013, 07:48, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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22 Mar 2006, 23:27
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lhotseface wrote:
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !

A. 38
B. 39
C. 40
D. 41
E. 42

350/7=50
77/7=11

so 50-11=39
or 50-2(11)=28
or 50-3(11)=17
or 50-4(11)=6

hence except 39 rest are to small than the answer choices. so go with B.
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22 Mar 2006, 23:35
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I'm getting 40

350 = 50 * 7
& 77 = 11 * 7

50*7 has 50 terms.

we can write it as,
39*7 + 77 = 350 => 40 terms.
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23 Mar 2006, 03:58
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lhotseface wrote:
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !

A. 38
B. 39
C. 40
D. 41
E. 42

Let x and y be the number of term 7 and the number of term 77 respectively. WE have :
7x+ 77y= 350 ---> x+11y = 50 ---> x+y= 50- 10y
we have : x+y= n = 50- 10y ---> n must have the unit digit of 0 because 50 - 10y must be a number which has unit digit of 0

Among the answer choices provided, only C matches.

I go for C.
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23 Mar 2006, 05:24
HIMALAYA wrote:
lhotseface wrote:
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !

A. 38
B. 39
C. 40
D. 41
E. 42

350/7=50
77/7=11

so 50-11=39
or 50-2(11)=28
or 50-3(11)=17
or 50-4(11)=6

hence except 39 rest are to small than the answer choices. so go with B.

You just missed the 77 to count
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Joined: 05 Apr 2005
Posts: 1710

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23 Mar 2006, 07:33
vivek123 wrote:
HIMALAYA wrote:
lhotseface wrote:
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !

A. 38
B. 39
C. 40
D. 41
E. 42

350/7=50
77/7=11

so 50-11=39
or 50-2(11)=28
or 50-3(11)=17
or 50-4(11)=6

hence except 39 rest are to small than the answer choices. so go with B.

You just missed the 77 to count

thats right. yah, i forget to count that.
Current Student
Joined: 29 Jan 2005
Posts: 5218

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25 Mar 2006, 22:52
vivek123 wrote:
I'm getting 40

350 = 50 * 7
& 77 = 11 * 7

50*7 has 50 terms.

we can write it as,
39*7 + 77 = 350 => 40 terms.

Worked it the same way as Vivek. Laxie`s method would be more efficient with much bigger numbers...
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Joined: 02 Sep 2010
Posts: 803
Location: London

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19 Oct 2010, 00:50
Merging topics ...

Please try using the forum search feature ... You should be able to find answers to a lot of questions which have been asked before
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19 Oct 2010, 14:02
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satishreddy wrote:
if each term in the sum of 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

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).

*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).

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.

Hope it helps.
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27 Jun 2011, 22:26
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Since, there is no 50 in the answer choices (350/7 = 50), we know there is at least one 77.

350 - 77 = 273
273/7 = 39
39+1 = 40.

If 40 wasn't there, I would have subtracted 77 from 273 and continued in a similar way.
Ans. C
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28 Jun 2011, 09:51
laxieqv wrote:
lhotseface wrote:
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !

A. 38
B. 39
C. 40
D. 41
E. 42

Let x and y be the number of term 7 and the number of term 77 respectively. WE have :
7x+ 77y= 350 ---> x+11y = 50 ---> x+y= 50- 10y
we have : x+y= n = 50- 10y ---> n must have the unit digit of 0 because 50 - 10y must be a number which has unit digit of 0

Among the answer choices provided, only C matches.

I go for C.

+1 Kudos to you laxieqv !!
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Joined: 01 Feb 2011
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29 Jun 2011, 17:15
x+11y = 50

x+y = n = verifying this with answer choices and checking for integer value for y, we can find out x+y = 40.

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29 Jun 2011, 17:29
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laxieqv wrote:
lhotseface wrote:
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !

A. 38
B. 39
C. 40
D. 41
E. 42

Let x and y be the number of term 7 and the number of term 77 respectively. WE have :
7x+ 77y= 350 ---> x+11y = 50 ---> x+y= 50- 10y
we have : x+y= n = 50- 10y ---> n must have the unit digit of 0 because 50 - 10y must be a number which has unit digit of 0

Among the answer choices provided, only C matches.

I go for C.

I like this method because it attacks the the question dead on by representing each term as x and y and then deductively finding the answer through POE. Hope I can think this quickly on the exam!
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Posts: 76
Concentration: Finance, Real Estate
WE: Asset Management (Real Estate)

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28 Jul 2011, 12:57
I used a different method altogether. I observed that since all the numbers are 7s, the total number of sevens has to be a multiple of 5 in order to get to 350. out of the answer choices, only 40 is a multiple of 5, So eliminate everything but choice C. I guess, if there were other multiples of 5 then you could use one of the methods above. But as a quick elimination strategy it helped to do the least amount of work.
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28 Jul 2011, 13:03
Substitution ..

it will take 50 7s to get 350 .. options are less than 50 .. so there should be atleast one 77.. 77/7 = 11 ..

so 39 7s + one 77 = 40
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29 Jul 2011, 13:43
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C it is..
350 = 7x+77y
x+11y=50

x y
39 1 =40
28 2
17 3
6 4

40 fits into our answer choice.

lhotseface wrote:
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !

A. 38
B. 39
C. 40
D. 41
E. 42
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30 Jul 2011, 21:58
+1 C
7(50)=350
Since 50 is not there in ANSWERS
so 77 needs to be ther
77=7(11)
So :
number of terms if one 77 is there 50-11+1=50-10=40 --> Check the Ans. C matches..
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30 Jul 2011, 22:06
A beautiful variant of this problem -
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350

Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
n has a value of :
I. 30
II. 45
III. 10
IV. 50
V. 25

Which of the following can be True ?
A - I,II,III
B - II,III,V
C - I,III,IV
D - All
C - None

Apply the same concepts and you can figure out the solution
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Re: If each term in the sum a1 + a2 + a3.....+ an is either 7 or [#permalink]

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30 Oct 2013, 04:23
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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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30 Oct 2013, 07:49
OPEN DISCUSSION OF THIS QUESTION IS HERE: if-each-term-in-the-sum-a1-a2-a3-an-is-either-7-or-93974.html
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Re: If each term in the sum a1+a2+a3+.....+an is either 7 or 77   [#permalink] 30 Oct 2013, 07:49
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