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18 Jan 2021, 07:30
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x + y: 12
m + n: 14
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
27% (01:47) correct
73%
(03:03)
wrong
based on 340
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\(x\), \(y\), \(m\), and \(n\) are positive integers. When \(x\) is divided by \(m\), the remainder is 8, and when \(y\) is divided by \(n\), the remainder is 4.
Select the least possible value of x + y for x + y, and select the least possible value of m + n for m + n. Make only two selections, one in each column.
Select the least possible value of x + y for x + y, and select the least possible value of m + n for m + n. Make only two selections, one in each column.
| x + y | m + n | |
| 12 | ||
| 13 | ||
| 14 | ||
| 15 | ||
| 16 | ||
| 26 |
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Official Answer
x + y: 12
m + n: 14
22 Nov 2024, 10:28
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Bunuel
Official Solution:
Theory: Positive integer \(a\) divided by positive integer \(d\) yielding a remainder of \(r\) can always be expressed as \(a = qd + r\), where \(q\) is the quotient and \(r\) is the remainder. Note here that \(0 \leq r < d\) (the remainder is a non-negative integer and always less than the divisor).
So, we’d have \(x = mq + 8\), from which it follows that \(x \geq 8\) (the dividend must be greater than or equal to the remainder), and \(m > 8\) (the divisor must be greater than the remainder). The least values of \(x\) and \(m\) are therefore 8 and 9, respectively. For example, \(x = 8\) divided by \(m = 9\) gives the remainder of 8.
Similarly, we’d have \(y = nq + 4\), from which it follows that \(y \geq 4\) (the dividend must be greater than or equal to the remainder), and \(n > 4\) (the divisor must be greater than the remainder). The least values of \(y\) and \(n\) are therefore 4 and 5, respectively. For example, \(y = 4\) divided by \(n = 5\) gives the remainder of 4.
Therefore, the least possible value of \(x + y\) is \(8 + 4 = 12\), and the least possible value of \(m + n\) is \(9 + 5 = 14\).
Correct answer:
\(x + y\) "12"
\(m + n\) "14"
Attachment:
GMAT-Club-Forum-f1zdy6nl.png [ 9.76 KiB | Viewed 1165 times ]
22 Nov 2024, 12:59
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1. We are given 4 positive integers - x, y, m, and n - and are asked to find the minimum value of the sums x + y and m + n. Note that m with n and x with y are unrelated, so finding the minimum sum is the same as finding the minimum of each number.
2. Let's consider m and n first. Both are positive integers and thus are \(\geq 0\). When something is divided by a number and given a remainder, the remainder must be less than that number. So, m > 8 and n > 4. The smallest they can be is 9 and 5, with a sum of 14.
3. Now let's consider x and y. Both are positive integers and thus are \(\geq 0\). To have a certain remainder they need to be greater or equal to it. So, x \(\geq\) 8 and y \(\geq\) 4. The smallest they can be is 8 and 4, with a sum of 12.
4. Our answer will be: x+y - 12 and m+n - 14.
2. Let's consider m and n first. Both are positive integers and thus are \(\geq 0\). When something is divided by a number and given a remainder, the remainder must be less than that number. So, m > 8 and n > 4. The smallest they can be is 9 and 5, with a sum of 14.
3. Now let's consider x and y. Both are positive integers and thus are \(\geq 0\). To have a certain remainder they need to be greater or equal to it. So, x \(\geq\) 8 and y \(\geq\) 4. The smallest they can be is 8 and 4, with a sum of 12.
4. Our answer will be: x+y - 12 and m+n - 14.
