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14 Jun 2022, 22:44
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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:
36% (02:28) correct
64%
(01:57)
wrong
based on 55
sessions
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If \(x\) and \(y\) are positive integers, what is the highest common divisor of \(x\) and \(y\)?
1) \(4x+3y=47\)
2) \(3x+4y=51\)
1) \(4x+3y=47\)
2) \(3x+4y=51\)
Updated on: 24 Oct 2023, 07:50
Originally posted by EthanTheTutor on 14 Jun 2022, 23:04.
Last edited by EthanTheTutor on 24 Oct 2023, 07:50, edited 1 time in total.
Last edited by EthanTheTutor on 24 Oct 2023, 07:50, edited 1 time in total.
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EthanTheTutor
Let \(F\) be the highest common divisor of \(x\) and \(y\).
Then both \(x\) and \(y\) are an integer multiple of \(F\), and we can write that:
\(x=mF\)
\(y=nF\)
Statement 1: \(4x+3y=47\)
⇒ \(4mF+3nF=47\)
⇒ \(F(4m+3n)=47\)
Since both \(F\) and \((4m+3n)\) are integers, they must both be divisors of \(47\).
The only divisors of \(47\) are \(1\) and \(47\), so either \(F=1\) or \(F=47\).
If \(F=47\), then the minimum possible values for \(x\) and \(y\) (since they are positive integers) are \(x=y=47\).
But \(4\cdot47+3\cdot47>47\) (not equal, as statement 1 requires), so \(F\ne47\).
Thus, \(F=1\). Sufficient
Statement 2: \(3x+4y=51\)
⇒ \(3mF+4nF=51\)
⇒ \(F(3m+4n)=51\)
Since both \(F\) and \((3m+4n)\) are integers, they must both be divisors of \(51\).
The divisors of \(51\) are \(1\), \(3\), \(17 \), and \(51\), so \(F\) is equal to either \(1\), \(3\), \(17 \), or \(51\).
If \(F=51\), then the minimum possible values for \(x\) and \(y\) (since they are positive integers) are \(x=y=51\).
But \(3\cdot51+4\cdot51>51\) (not equal, as statement 2 requires), so \(F\ne51\).
If \(F=17\), then the minimum possible values for \(x\) and \(y\) (since they are positive integers) are \(x=y=17\).
But \(3\cdot17+4\cdot17>17\) (not equal, as statement 2 requires), so \(F\ne17\).
If \(F=3\), then the minimum possible values for \(x\) and \(y\) (since they are positive integers) are \(x=y=3\).
Since \(3\cdot3+4\cdot3<51\) we cannot immediately rule out \(3\) as the highest common divisor of \(x\) and \(y\).
By testing values (it helps to recognize that \(y\) must be a multiple of \(3\), and \(x\) must NOT be a multiple of \(2\)), we can find the following set of solutions:
\(y=3\) and \(x=13\) ... \(HCD=1\)
\(y=6\) and \(x=9\) ... \(HCD=3\)
\(y=9\) and \(x=5\) ... \(HCD=1\)
\(y=12\) and \(x=1\) ... \(HCD=1\)
This demonstrates that the highest common divisor of \(x\) and \(y\) can be either \(1\) or \(3\). Not Sufficient
24 Oct 2023, 06:11
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