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08 Nov 2023, 12:06
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When 85 is divided by a positive integer \(n\), the remainder is 7, and when 85 is divided by \(2n\), the remainder is \(n + 7\). How many different values can \(n\) take?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
08 Nov 2023, 12:06
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Official Solution:
When 85 is divided by a positive integer \(n\), the remainder is 7, and when 85 is divided by \(2n\), the remainder is \(n + 7\). How many different values can \(n\) take?
A. 0
B. 1
C. 2
D. 3
E. 4
Dividing 85 by \(n\), and obtaining a remainder of 7, can be expressed as \(85 = nq + 7\).
Similarly, dividing 85 by \(2n\), and obtaining a remainder of \(n+7\), can be expressed as \(85 = 2np + (n+7)\).
The first equation, \(85 = nq + 7\), simplifies to \(nq = 78\), indicating that \(n\) is a factor of 78. The factors of 78 are 1, 2, 3, 6, 13, 26, 39, and 78. However, since \(n\) must be greater than 7 (because the divisor is always larger than the remainder), the possible values for \(n\) are 13, 26, 39, or 78.
The second equation, \(85 = 2np + (n+7)\), implies that \(n+7\) must be odd because \(2np\) is even and their sum must equal the odd number 85. Thus, \(n\) must be even. This narrows the possible values of \(n\) to 26 and 78.
For \(n=26\), dividing 85 by \(2*26 = 52\) leaves a remainder of 33, which is equal to \(n+7=26+7\). Hence, this value meets the conditions.
For \(n=78\), dividing 85 by \(2*78 = 156\) leaves a remainder of 85, which is equal to \(n+7=78+7\). Hence, this value meets the conditions.
If \(n=78\), then 85 divided by \(2*78=156\) gives the reminder of 85, which does equal to \(n+7=78+7\). So, this value satisfies the conditions.
Therefore, \(n\) can take two values: 26 and 78.
Answer: C
When 85 is divided by a positive integer \(n\), the remainder is 7, and when 85 is divided by \(2n\), the remainder is \(n + 7\). How many different values can \(n\) take?
A. 0
B. 1
C. 2
D. 3
E. 4
Dividing 85 by \(n\), and obtaining a remainder of 7, can be expressed as \(85 = nq + 7\).
Similarly, dividing 85 by \(2n\), and obtaining a remainder of \(n+7\), can be expressed as \(85 = 2np + (n+7)\).
The first equation, \(85 = nq + 7\), simplifies to \(nq = 78\), indicating that \(n\) is a factor of 78. The factors of 78 are 1, 2, 3, 6, 13, 26, 39, and 78. However, since \(n\) must be greater than 7 (because the divisor is always larger than the remainder), the possible values for \(n\) are 13, 26, 39, or 78.
The second equation, \(85 = 2np + (n+7)\), implies that \(n+7\) must be odd because \(2np\) is even and their sum must equal the odd number 85. Thus, \(n\) must be even. This narrows the possible values of \(n\) to 26 and 78.
For \(n=26\), dividing 85 by \(2*26 = 52\) leaves a remainder of 33, which is equal to \(n+7=26+7\). Hence, this value meets the conditions.
For \(n=78\), dividing 85 by \(2*78 = 156\) leaves a remainder of 85, which is equal to \(n+7=78+7\). Hence, this value meets the conditions.
If \(n=78\), then 85 divided by \(2*78=156\) gives the reminder of 85, which does equal to \(n+7=78+7\). So, this value satisfies the conditions.
Therefore, \(n\) can take two values: 26 and 78.
Answer: C
01 Jan 2024, 10:45
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Hi, I'm confused by 78 as one of the values. Isnt the remainder of 85 divided by 156 70, not 85?
Also, if we know that n is even, why do we not test 2 and 6 as well, to verify that these 2 options should be eliminated?
Thank you in advance for any clarification!
Also, if we know that n is even, why do we not test 2 and 6 as well, to verify that these 2 options should be eliminated?
Thank you in advance for any clarification!
