Question

A number of two digits exceeds twice the sum of its digits by 26. If the
digits are interchanged, the resulting number exceeds the given number
by 18. Find the number.

Answer 1

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Answer 2

Given:

A 2-digit number that exceeds twice the sum of its digits by 26.

On interchanging, the resulting number exceeds given number by 18.

To find:

The value of the 2-digit number.

Solution:

Let the ones place of the 2-digit number be $y$ and the tens place be $x$. Hence the 2-digit number will be $10x+y$. On interchanging digits, the number will be $10y+x$.

From the given information,

$10x+y=2(x+y)+26$

$10x+y=2x+2y+26$

$10x-2x=2y-y+26$

$8x-y=26...(1)$

On interchanging,

$10y+x=10x+y+18$

$10y-y=10x-x+18$

$9y=9x+18$

$y-x=2...(2)$

Adding equations (1) and (2)

$-y+8x=26$

$y-x=2$

$7x=28$

$x=\frac{28}{7}=4$

Substituting in equation (2)

$y-x=2$

$y-4=2$

$y=4+2=6$

Hence, the number is:

$10x+y=10(4)+7=40+7=47$

On interchanging, it becomes $74$.

The number is calculated as 47.