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

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(C) Let the digit at the ten's place be $y$ and the digit at the unit's place be $x$.The original number is $10y + x$.According to the problem,the sum

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(C) Let the digit at the ten's place be $y$ and the digit at the unit's place be $x$.The original number is $10y + x$.According to the problem,the sum of the digits is $6$:$x + y = 6$  ............ $(1)$When the digits are interchanged,the new number is $10x + y$.According to the problem,the new number is $18$ less than the original number:$(10y + x) - (10x + y) = 18$$9y - 9x = 18$Dividing by $9$,we get:$y - x = 2$  ............ $(2)$Adding equation $(1)$ and equation $(2)$:$(x + y) + (y - x) = 6 + 2$$2y = 8$$y = 4$Substituting $y = 4$ into equation $(1)$:$x + 4 = 6$$x = 2$Therefore,the original number is $10(4) + 2 = 42$.

The sum of the digits of a two-digit number is $6$. The number obtained by interchanging the digits is $18$ less than the original number. Find the original number.

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