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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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)