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When positive integer \(n\) is divided by 3, the remainder

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by swerve » Tue May 21, 2019 12:12 pm

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When positive integer \(n\) is divided by 3, the remainder is 1. When \(n\) is divided by 7, the remainder is 5. What is the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21?

A. 2
B. 2
C. 5
D. 19
E. 20

The OA is B

Source: Veritas Prep
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by Jay@ManhattanReview » Tue May 21, 2019 10:00 pm
swerve wrote:When positive integer \(n\) is divided by 3, the remainder is 1. When \(n\) is divided by 7, the remainder is 5. What is the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21?

A. 1
B. 2
C. 5
D. 19
E. 20

The OA is B

Source: Veritas Prep
Given that the positive integer \(n\) is divided by 3, and the remainder is 1, we have n, one among 4, 7, 13, 16, 19, 22 ...

Also, given that the positive integer \(n\) is divided by 7, and the remainder is 5, we have n, one among 12, 19, 26, ...

The smallest common value for n is 19.

Since n = 19 divided by 21 leaves a remainder of 19, for it to be divisible by 21, we need to add 2 to it. Thus, the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21 is 2.

The correct answer: B

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Thu May 23, 2019 4:12 pm
swerve wrote:When positive integer \(n\) is divided by 3, the remainder is 1. When \(n\) is divided by 7, the remainder is 5. What is the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21?

A. 2
B. 2
C. 5
D. 19
E. 20

The OA is B

Source: Veritas Prep
Since when n is divided by 3, the remainder is 1, n could be:

1, 4, 7, 10, 13, 16, 19, 22, ...

Since when n is divided by 7, the remainder is 5, n could be:

5, 12, 19, 26, ...

We can see that n = 19 satisfies both division/remainder criteria. And if p = 2, we have n + p = 19 + 2 = 21, which is a multiple of 21.

Alternate Solution:

Since the remainder from the division of n by 3 is 1, we can express n as n = 3k + 1 for some integer k.

Similarly, n = 7s + 5 for some integer s.

We see that n + 2 = 3k + 3 = 7s + 7 is divisible by both 3 and 7; therefore it must be divisible by 21 as well. So, the smallest such integer is 2.

Answer: A and B (Note the multiple 2 among answer choices)

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