The sum of two numbers is 468 and their HCF i | vedclass

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(B) Let the two numbers be $39x$ and $39y$,where $x$ and $y$ are coprime (i.e.,$gcd(x, y) = 1$).Given that the sum of the numbers is $468$,we have $39

Vedclass · Accepted Answer

$2$

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(B) Let the two numbers be $39x$ and $39y$,where $x$ and $y$ are coprime (i.e.,$gcd(x, y) = 1$).Given that the sum of the numbers is $468$,we have $39x + 39y = 468$.Dividing by $39$,we get $x + y = \frac{468}{39} = 12$.We need to find pairs of coprime numbers $(x, y)$ such that $x + y = 12$ and $x The possible pairs $(x, y)$ that sum to $12$ are:$(1, 11)$ where $gcd(1, 11) = 1$ (Valid)$(2, 10)$ where $gcd(2, 10) = 2$ (Invalid)$(3, 9)$ where $gcd(3, 9) = 3$ (Invalid)$(4, 8)$ where $gcd(4, 8) = 4$ (Invalid)$(5, 7)$ where $gcd(5, 7) = 1$ (Valid)$(6, 6)$ where $gcd(6, 6) = 6$ (Invalid)Thus,there are $2$ such pairs: $(1, 11)$ and $(5, 7)$.

The sum of two numbers is $468$ and their $HCF$ is $39$. How many pairs of such numbers can be formed?

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