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Solution:

Let there be a number of 7s and b number of 77s

=> Total number of terms = a + b

=>7a + 77b = 350

=> a + 11b = 50

Its a Diophantine equation and we can solve it by checking for values of a and b

If b=1 then a should be 39 and total terms = a +b =1+39 =40

=> Total terms is present among the options

Hence 39 (option c)

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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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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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Bunuel
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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Bunuel
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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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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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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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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LM
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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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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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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Amazing question!
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Best method:

Since all the numbers have unit digit 7 so all we need to think is how many 7's added together result in a number with unit digit 0 (unit of 350)

and the answer is 7 must be added 10 time or multiple of 10 times


The only option that is multiple of 10 here is C hence

Correct answer : Option C
:)

LM
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
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