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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.