Question

A number has three digits. The sum of its digits is equal to of the number if 99 is added to the
number, then digits at the extremes are interchanged. If the middle digit is three times the sum of
other two, find the number. [Number = 192]​

Answer 1

Let our three digit number be $100x+10y+z.$

Here the sum of the digits of the number is equal to $\left(\dfrac{1}{16}\right)^{th}$  of the number, i.e.,

$\longrightarrow \dfrac{100x+10y+z}{16}=x+y+z$

$\longrightarrow 100x+10y+z=16x+16y+16z\quad\quad\dots(1)$

Given, our number gets reversed if we add 99 to it, i.e.,

$\longrightarrow 100z+10y+x=100x+10y+z+99$

$\longrightarrow 100z+10y+x-100x-10y-z=99$

$\longrightarrow 99z-99x=99$

$\longrightarrow 99(z-x)=99$

$\longrightarrow z-x=1$

$\longrightarrow z=x+1\quad\quad\dots(2)$

Putting (2) in (1),

$\longrightarrow 100x+10y+x+1=16x+16y+16(x+1)$

$\longrightarrow 101x+10y+1=16x+16y+16x+16$

$\longrightarrow 101x+10y+1=32x+16y+16$

$\longrightarrow 69x-6y-15=0\quad\quad\dots(3)$

But given that, the middle digit of the number is three times the sum of the other two numbers, i.e.,

$\longrightarrow y=3(x+z)$

From (2),

$\longrightarrow y=3(x+x+1)$

$\longrightarrow y=3(2x+1)$

$\longrightarrow y=6x+3\quad\quad\dots(4)$

So, putting (4) in (3),

$\longrightarrow 69x-6(6x+3)-15=0$

$\longrightarrow 69x-36x-18-15=0$

$\longrightarrow 33x-33=0$

$\longrightarrow 33x=33$

$\longrightarrow x=1$

From (4),

$\longrightarrow y=6\times1+3$

$\longrightarrow y=9$

From (2),

$\longrightarrow z=1+1$

$\longrightarrow z=2$

Hence our three digit number is,

$\longrightarrow 100x+10y+z=100\times1+10\times9+2$

$\longrightarrow\underline{\underline{100x+10y+z=192}}$