A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Given:

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. 

To do:

We have to find the number.

Solution:

Let the two-digit number be $10x+y$.

Sum of the digits $=x+y$.

Difference of the digits $=x-y$ or $y-x$.

According to the question,
$10x+y=8\times(x+y)-5$
$10x+y=8x+8y-5$

$10x-8x=8y-y-5$

$2x=7y-5$ .....(i)

$10x+y=16\times(x-y)+3$ or $10x+y=16\times(y-x)+3$

$10x+y=16x-16y+3$ or $10x+y=16y-16x+3$

$16x-10x+3=y+16y$ or $10x+16x=16y-y+3$

$6x+3=17y$ or $26x=15y+3$

$3(2x)+3=17y$ or $13(2x)=15y+3$

$3(7y-5)+3=17y$ or $13(7y-5)=15y+3$     (From (i))

$21y-15+3=17y$ or $91y-65=15y+3$

$21y-17y-12=0$ or $91y-15y=3+65$

$4y=12$ or $76y=68$

$y=\frac{12}{4}$ or $y=\frac{68}{76}$

$y$ cannot be a fraction. Therefore,

$y=3$

Substituting $y=3$ in equation (i), we get,

$2x=7(3)-5$

$2x=21-5$

$2x=16$

$x=\frac{16}{2}$

$x=8$

The number is $10x+y=10(8)+3=80+3=83$.

The required number is 83.

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Updated on: 10-Oct-2022

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