The difference between a two-digit number and | vedclass

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(B) Let the two-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.According to the problem,the difference between the n

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$8$

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(B) Let the two-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.According to the problem,the difference between the number and the number obtained by interchanging its digits is $36$:$| (10x + y) - (10y + x) | = 36$$| 9x - 9y | = 36$$| x - y | = 4$Given the ratio of the digits is $1:2$,we have $y/x = 1/2$ or $x/y = 1/2$. Thus,$x = 2y$ or $y = 2x$.Substituting $x = 2y$ into $| x - y | = 4$,we get $| 2y - y | = 4$,which means $y = 4$. Then $x = 2(4) = 8$.The digits are $8$ and $4$.The sum of the digits is $x + y = 8 + 4 = 12$.The difference of the digits is $x - y = 8 - 4 = 4$.The difference between the sum and the difference of the digits is $12 - 4 = 8$.

The difference between a two-digit number and the number obtained by interchanging the digits is $36$. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is $1:2$?

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