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

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

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New post Updated on: 26 Feb 2019, 19:58
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A
B
C
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  35% (medium)

Question Stats:

83% (01:36) correct 17% (02:34) wrong based on 6 sessions

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How to solve this?

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

Originally posted by prottenbucher on 26 Feb 2019, 16:00.
Last edited by u1983 on 26 Feb 2019, 19:58, edited 1 time in total.
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Re: If each term in the sum a1 + a2 + . . . + an is either 7 or 77...  [#permalink]

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New post 26 Feb 2019, 20:00
If all 7, n =50 but then there are no 77.
If there is 1 77 , then that 77 accounts for 11 7s.
Hence total no of terms 39 (7s) +1 (77) = 40
Ans C
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If each term in the sum a1 + a2 + . . . + an is either 7 or 77...  [#permalink]

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New post 26 Feb 2019, 20:04
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Let’s say that there are x terms of 7 and y terms of 77 in the sum.
So, 7*x+77*y=350.
7(x+11y)=350
x+11y=50
Now, there is a limited number of values that y can take: 1, 2, 3, 4.
If y=1, then 39+11*1=50 --- x+y=40 (we have this answer choice).
If y=2, then 28+11*2=50 --- x+y=30 (we don’t have this answer choice).
If y=3, then 17+11*3=50 --- x+y=20 (we don’t have this answer choice).
If y=4, then 6+11*4=50 --- x+y=10 (we don’t have this answer choice).

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

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New post 26 Feb 2019, 20:31
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a1+a2+a3......an = 350.
since the terms can be either 7 or 77 we take 7 common
7(a+11b) = 350
=> a + 11b = 50, where a is the number of terms equal to 7 and b is the number of terms equal to 77
Since the options are between 42 to 38, we know that b > 0.
(if b = 0, it would take 1*50 = 50 terms to get the sum 350)

take b=1
=> 50 - 11 = 39
1*39 = 39 (a=39)
Therefore number of terms = a+b = 1+ 39 = 40 (answer)
Increasing b would reduce the total number of terms below 38
check with b = 2
=> 50 - 22 = 28
=> Number of terms = 28 + 1 = 29(which is beyond the range of options). Therefore, we do not need to check the others
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Re: If each term in the sum a1 + a2 + . . . + an is either 7 or 77...  [#permalink]

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New post 26 Feb 2019, 21:12
1
prottenbucher wrote:
How to solve this?

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


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Re: If each term in the sum a1 + a2 + . . . + an is either 7 or 77...   [#permalink] 26 Feb 2019, 21:12
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