Question

this sum of the digits of a two-digit number is equal to 1 by 8 of the number if the digits of the numbers are reversed the new number is 45 less than the original number find the original number

Answer 1

Hi ,

let ten's place digit = x

unit place digit = y


the original number = 10x + y -----( 1 )

according to the problem given,


sum of the digits = 1 / 8 of the number

x + y = 1/8( 10x + y )
8 ( x + y ) = 10x + y

8x + 8y = 10x + y

8y -y = 10x - 8x

7y = 2x

7y/ 2 = x

x = 7y / 2 -----( 2 )

if the digits of the number reversed the

new number formed = 10y + x ---( 3 )

( 1 ) - ( 2 ) = 45

10x + y - ( 10y + x ) = 45

10x + y - 10y - x = 45

9x - 9y = 45

divide each term with 9 , we get

x - y = 5

x = y + 5-----( 4 )

equation ( 2 ) = equation ( 4 )

7y/2 = y + 5


7y = 2 ( y + 5 )

7y = 2y + 10

7y - 2y = 10

5y = 10

y = 10 / 5

y = 2

put y = 2 in equation ( 4 ), we get

x = 2 + 5

x = 7

Therefore ,

x = 7 , y = 2

the original number = 10x + y

= 10× 7 + 2

= 70 + 2

= 72

I hope this helps you.

:)