A two-digit number is obtained by either mult | 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 first condition: $10x + y = 8(x + y)

Vedclass · Accepted Answer

$41$

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 first condition: $10x + y = 8(x + y) + 1$.$10x + y = 8x + 8y + 1 \implies 2x - 7y = 1$  --- (Equation $1$).According to the second condition: $10x + y = 13(x - y) + 2$.$10x + y = 13x - 13y + 2 \implies -3x + 14y = 2$  --- (Equation $2$).Multiply Equation $1$ by $2$: $4x - 14y = 2$  --- (Equation $3$).Add Equation $2$ and Equation $3$: $(-3x + 14y) + (4x - 14y) = 2 + 2$.$x = 4$.Substitute $x = 4$ into Equation $1$: $2(4) - 7y = 1 \implies 8 - 7y = 1 \implies 7y = 7 \implies y = 1$.The number is $10(4) + 1 = 41$.

$A$ two-digit number is obtained by either multiplying the sum of the digits by $8$ and adding $1$,or by multiplying the difference of the digits by $13$ and adding $2$. Find the number.

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