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n ranges over the positive integers between 100 and 200, inclusive. Fi

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Math Revolution GMAT Instructor
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n ranges over the positive integers between 100 and 200, inclusive. Fi  [#permalink]

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New post 31 Jul 2019, 00:09
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Question Stats:

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[GMAT math practice question]

\(n\) ranges over the positive integers between \(100\) and \(200,\) inclusive. Find the number of values of \(7n+2\) which are multiples of \(5\)

\(A. 18\)

\(B. 20\)

\(C. 22\)

\(D. 24\)

\(E. 26\)

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Re: n ranges over the positive integers between 100 and 200, inclusive. Fi  [#permalink]

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New post 31 Jul 2019, 02:52
7n+2= 0 mod 5
7n= -2 mod 5= 3 mod 5
n= 4 mod 5

104= 4 mod 5
.
.
199= 4 mod 5

total possible values of n= [(199-104)/5]+1=20




MathRevolution wrote:
[GMAT math practice question]

\(n\) ranges over the positive integers between \(100\) and \(200,\) inclusive. Find the number of values of \(7n+2\) which are multiples of \(5\)

\(A. 18\)

\(B. 20\)

\(C. 22\)

\(D. 24\)

\(E. 26\)
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Joined: 15 Feb 2019
Posts: 4
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Re: n ranges over the positive integers between 100 and 200, inclusive. Fi  [#permalink]

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New post 01 Aug 2019, 02:27
3
Hi,
I'm not sure if my approach is correct but here's how I did it.
7n+2 will be divisible by 5 if 7n has last digit as 3 or 8. There are only two digits (103 and 108) from 100 to 110. So from 100 to 200, there will be 2*10 digits which satisfy this condition.
Please correct me if i'm wrong. Would request your input Bunuel .
Cheers!
Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
Posts: 7766
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: n ranges over the positive integers between 100 and 200, inclusive. Fi  [#permalink]

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New post 02 Aug 2019, 02:02
=>

If \(n = 5k,\) then \(7n + 2 = 7(5k) + 2 = 5(7k) + 2\) is not a multiple of \(5.\)

If \(n = 5k+1,\) then \(7n + 2 = 7(5k+1) + 2 = 5(7k) + 7 + 2 = 5(7k) + 9 = 5(7k+1)+4\) is not a multiple of \(5.\)

If \(n = 5k+2,\) then \(7n + 2 = 7(5k+2) + 2 = 5(7k) + 14 + 2 = 5(7k) + 16 = 5(7k+3)+1\) is not a multiple of \(5.\)

If \(n = 5k+3,\) then \(7n + 2 = 7(5k+3) + 2 = 5(7k) + 21 + 2 = 5(7k) + 23 = 5(7k+4)+3\) is not a multiple of \(5.\)

If \(n = 5k+4,\) then \(7n + 2 = 7(5k+4) + 2 = 5(7k) + 28 + 2 = 5(7k) + 30 = 5(7k+6)\) is a multiple of \(5.\)

Thus, \(n\) has remainder \(4\) when it is divided by \(5.\)

The possible values of \(n\) are \(104, 109, …, 199.\)

The number of possible values of \(n\) is \(\frac{(199-104)}{5} + 1 = 20.\)

Therefore, B is the answer.
Answer: B
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Re: n ranges over the positive integers between 100 and 200, inclusive. Fi   [#permalink] 02 Aug 2019, 02:02
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