The sum of the digits of a two-digit number i | vedclass

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(B) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.According to the problem,the sum of the digits is $x + y = 13$ (Equa

Vedclass · Accepted Answer

$67$

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(B) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.According to the problem,the sum of the digits is $x + y = 13$ (Equation $1$).The number obtained by interchanging the digits is $10y + x$.According to the problem,$(10y + x) = (10x + y) + 9$.Simplifying this,$9y - 9x = 9$,which gives $y - x = 1$ (Equation $2$).Adding Equation $1$ and Equation $2$: $(x + y) + (y - x) = 13 + 1$,so $2y = 14$,which means $y = 7$.Substituting $y = 7$ into Equation $1$: $x + 7 = 13$,so $x = 6$.The original number is $10x + y = 10(6) + 7 = 67$.

The sum of the digits of a two-digit number is $13$. The number obtained by interchanging the digits is $9$ more than the original number. Find the original number.

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