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02 Sep 2020, 20:00
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D
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41% (02:11) correct
59%
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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.
M36-104
(1) The product of any two factors of k is odd
(2) The sum of any two factors of k is even.
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M36-104
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15 Nov 2021, 03:07
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Bunuel
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
General Discussion
02 Sep 2020, 20:18
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Asked: If k is a positive integer and 175 divided by k leaves a remainder of 7, what is the value of k?
175 =km + 7; where m is an integer
168 = km = 2^3×3×7
Since division by k, leaves remainder 7; k>7
(1) The product of any two factors of k is odd
k is odd; k>7 and km=168
k =21 is the only solution since k is odd, k>7 and I is a factor of 168.
SUFFICIENT
(2) The sum of any two factors of k is even.
k is odd since sum of 2 odd numbers is even, k cannot be even since 1 is a factor of k and is odd.
k is odd; k>7 and km=168
k =21 is the only solution since k is odd, k>7 and I is a factor of 168.
SUFFICIENT
IMO D
Posted from my mobile device
175 =km + 7; where m is an integer
168 = km = 2^3×3×7
Since division by k, leaves remainder 7; k>7
(1) The product of any two factors of k is odd
k is odd; k>7 and km=168
k =21 is the only solution since k is odd, k>7 and I is a factor of 168.
SUFFICIENT
(2) The sum of any two factors of k is even.
k is odd since sum of 2 odd numbers is even, k cannot be even since 1 is a factor of k and is odd.
k is odd; k>7 and km=168
k =21 is the only solution since k is odd, k>7 and I is a factor of 168.
SUFFICIENT
IMO D
Posted from my mobile device
