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



To get a ZERO at the end, 350, of a number, the some of 7s must at least be 10, because 7 + 7 + .. + 7 = 70 or 7*10

Knowing that we must have K*10 * 7, K = integer.
The only option is C.
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
Hi what is the level of this question...is it sub 600 or 600-700 or above 700 level

Posted from my mobile device
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
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RebelWanderer wrote:
Hi what is the level of this question...is it sub 600 or 600-700 or above 700 level

Posted from my mobile device


You can check the difficulty of a question among the tags just above the original post:



Hope it helps.
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
Bunuel wrote:
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

Number of approaches are possible.

For example, approach #1:

Since the units digit of 350 is zero then the number 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.

Approach #2:

\(7x+77y=350\), 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.




3+3+3+3+4+4 = 20 But here the number of terms is not a multiple of 10. Is there any specific thing I am missing can you help in the mentioned approach. Bunuel
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
Expert Reply
himgkp1989 wrote:
Bunuel wrote:
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

Number of approaches are possible.

For example, approach #1:

Since the units digit of 350 is zero then the number 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.

Approach #2:

\(7x+77y=350\), 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.




3+3+3+3+4+4 = 20 But here the number of terms is not a multiple of 10. Is there any specific thing I am missing can you help in the mentioned approach. Bunuel


You are missing that the units digits of all the terms must be the same, as it is in the question at hand.
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
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Given that If each term in the sum \(a_1+a_2+a_3+...+a_n\) is either 7 or 77 and the sum equals 350 and we need to find which of the following could be equal to n

Let's start by assuming that all numbers are 7
As there are n numbers so their sum will be 7*n = 7n = 350 (given)
=> n = \(\frac{350}{7}\) = 50

But this is not an option, so lets start relacing 7's by 77 and see what we get
To insert one 77 we need to replace 11 7's as 77 = 7*11

=> We will have 50-11 = 39 7's and one 77
=> Total number of terms = 39 + 1 = 40

So, Answer will be C
Hope it helps!

Watch the following video to learn How to Sequence problems

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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
Bunuel
Can you please let me know why cant i take y = 2 in x + 11y = 50 then x = 28 and 28+ 22 = 50 or y = 3 then x = 17
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
7*50=350

7*11=77 this is one term (suposse there is only one "77")

50-11=39, so:

77 + 7*39 = 350

One "77" plus 39 times 7 = 40 Terms
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
LM wrote:
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


77x+7y=350

(1)11x+y=50
(2) x+y=n /-1

10x=50-n
n=50-10x

x could be 1,2,3,4 or 5.

x=1->n=40
x=2->n=30
x=3->n=20
x=4->n=10
x=5->n=0

so n=40
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
Why couldn't x = 40 in your second example Bunuel? I'm trying to understand what made you choose x =39 in the first place.

Thank you!
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
This question can be simplified in the form of special equation as
7a + 77b= 350
a+11b=50
put b=1, now a=39 and therefore sum=40
put b=2, now a=28 and therefore sum=30
put b=3, now a=17 and therefore sum=20
put b=4, now a=6 and therefore sum=10

among the options only 40 matches henceforth the answer.
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum [#permalink]
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