Question

A number consists of three digits, the right-hand digit being zero. If the left hand and the middle by interchanged the number is diminished by $180$. If the left hand digits be halved and the middle and right-hand digits be incharged the number is diminished by $454$. Find the number.

Answer 1

Since the number contains three digits, it has a hundreds place, a tens place and a ones place.

It is given that the right most digit is $0$. Let the digits in the hundreds place and tens place be $x$ and $y$ respectively. Thus, the number is

$100x+10y+0=100x+10y.$

It is also given that if the digits in the hundreds place and tens place are interchanged, the difference between the original number and the new number after interchange is $180.$

$\Rightarrow 100x+10y-(100y+10x)=180$

$\Rightarrow 100x+10y-100y-10x=180$

$\Rightarrow 90x-90y=180$

$\Rightarrow 90(x-y)=180$

$\Rightarrow x-y=2$ ......(i)

It is also given that if the digit in the hundreds place is halved and digits in tens place and ones place are interchanged, the difference between the original number and the new number after interchange is $454.$

$\Rightarrow 100x+10y-(\frac{100x}{2}+0+y)=454$

$\Rightarrow 100x+10y-50x-y=454$

$\Rightarrow 50x+9y=454$ .....(ii)

Multiplying both sides of Eqn (i) by $50$, we get

$50x-50y=100$...(iii)

Eqn (i) $-$ Eqn (iii) we get,

$\Rightarrow (50x-50y)-(50x+9y)=100-454$

$\Rightarrow 50x-50y-50x-9y=-354$

$\Rightarrow -59y=-354$

$\Rightarrow y=6$

Now, substitute the value of $y$ in the equation $x-y=2.$

$\Rightarrow x-6=2$

$\Rightarrow x=8$

Therefore, the three digit number $xy0=860.$