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

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Question

(A) 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 sum of the digits is $x

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) 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 sum of the digits is $x + y = 8$ $...(1)$.When the digits are reversed,the new number is $10y + x$.Given that the new number is $54$ more than the original number: $10y + x = (10x + y) + 54$.Simplifying this equation: $9y - 9x = 54$,which gives $y - x = 6$ $...(2)$.Adding equations $(1)$ and $(2)$: $(x + y) + (y - x) = 8 + 6$,so $2y = 14$,which means $y = 7$.Substituting $y = 7$ into equation $(1)$: $x + 7 = 8$,so $x = 1$.The original number is $10x + y = 10(1) + 7 = 17$.

The sum of the digits of a two-digit number is $8$. If the digits are reversed,the number is increased by $54$. Find the number.

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