The difference between a two-digit number and the number formed by reversing its digits is 18. If the sum of the digits of the original number is 8, then find the number.
Answers
Let x be the digit at unit’s place and y be the digit at ten’s place.
Given:
Since y is at ten’s place, then the number formed is 10y+x.
By reversing the digits, it becomes 10x+y.
As the difference of the numbers is 18, so,
(10y+x)−(10x+y)=18
9(y−x)=18
y−x=2 -------(1)
As the sum of digits is 8, so,
x+y=8 -------(2)
On adding equations (1) and (2), we get
y−x=2
+ y+x=8
2y =10
⇒y=5
Putting this in (2), we get
x+y=8
x+5=8
x=8-5
⇒x=3
Therefore,
⇒x=3,y=5
Hence, number =10y+x
=10×5+3
=53.
Hope you understand
Answer:
53
Step-by-step explanation:
Let x be the digit at unit’s place and y be the digit at ten’s place.
Given:
Since y is at ten’s place, then the number formed is 10y+x.
By reversing the digits, it becomes 10x+y.
As the difference of the numbers is 18, so,
(10y+x)−(10x+y)=18
9(y−x)=18
y−x=2 -------(1)
As the sum of digits is 8, so,
x+y=8 -------(2)
On adding equations (1) and (2), we get
y−x=2
+ y+x=8
2y =10
⇒y=5
Putting this in (2), we get
x+y=8
x+5=8
x=8-5
⇒x=3
Therefore,
⇒x=3,y=5
Hence, number =10y+x
=10×5+3
=53.
Hope you understand