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If \(k\) is a positive integer and 175 divided by \(k\) leaves a remainder of 7, what is the value of \(k\)?
(1) The product of any two factors of \(k\) is odd.
(2) The sum of any two factors of \(k\) is even.
(1) The product of any two factors of \(k\) is odd.
(2) The sum of any two factors of \(k\) is even.
15 Nov 2021, 03:07
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Official Solution:
If \(k\) is a positive integer and 175 divided by \(k\) leaves a remainder of 7, what is the value of \(k\)?
175 divided by \(k\) leaves a remainder of 7: \(175=kq + 7\) (notice here that \(k\) (the divisor) must be greater than 7 (the remainder)):
\(168=kq\);
\(kq=2^3*3*7\).
(1) The product of any two factors of \(k\) is odd.
\(k\) cannot be even because if it is then it has 1 and 2 as factors and \(1*2=2=even\). Thus \(k\) must be odd. In this case all its factors will be odd and the product of any two distinct positive factors of \(k\) will be odd. Now, if \(k=odd\) and \(k > 7\), then from \(kq=2^3*3*7\) it follows that \(k=21\) (the only odd factor of 168 greater than 7). Sufficient.
(2) The sum of any two factors of \(k\) is even.
\(k\) cannot be even because if it is then it has 1 and 2 as factors and \(1*2=2=even\). Thus \(k\) must be odd. In this case all its factors will be odd and the sum of any two distinct positive factors of \(k\) will be even. Now, if \(k=odd\) and \(k > 7\), then from \(kq=2^3*3*7\) it follows that \(k=21\) (the only odd factor of 168 greater than 7). Sufficient.
Answer: D
If \(k\) is a positive integer and 175 divided by \(k\) leaves a remainder of 7, what is the value of \(k\)?
175 divided by \(k\) leaves a remainder of 7: \(175=kq + 7\) (notice here that \(k\) (the divisor) must be greater than 7 (the remainder)):
\(168=kq\);
\(kq=2^3*3*7\).
(1) The product of any two factors of \(k\) is odd.
\(k\) cannot be even because if it is then it has 1 and 2 as factors and \(1*2=2=even\). Thus \(k\) must be odd. In this case all its factors will be odd and the product of any two distinct positive factors of \(k\) will be odd. Now, if \(k=odd\) and \(k > 7\), then from \(kq=2^3*3*7\) it follows that \(k=21\) (the only odd factor of 168 greater than 7). Sufficient.
(2) The sum of any two factors of \(k\) is even.
\(k\) cannot be even because if it is then it has 1 and 2 as factors and \(1*2=2=even\). Thus \(k\) must be odd. In this case all its factors will be odd and the sum of any two distinct positive factors of \(k\) will be even. Now, if \(k=odd\) and \(k > 7\), then from \(kq=2^3*3*7\) it follows that \(k=21\) (the only odd factor of 168 greater than 7). Sufficient.
Answer: D
