Question

The sum of the digits of a two-digit number is 8. If the digits are reversed the number formed is 18 more than the original number. What is the original number?

Answer 1

Let the ten's digit and the one's digit be x and y respectively.

Given

The sum of the digits of a two-digit number is 8.

$\implies x+y=8--(1)$

If the digits are reversed the number formed is 18 more than the original number.

$\implies 10y+x=10x+y+18\\\implies -10x+x+10y-y=18\\\implies -9x+9y=18\\\implies -x+y=2--(2)$

Solving (1) and (2), we get,

$x+y=8\\\underline{-x+y=2}\\\underline{\underline{2y=10}}\\\implies y =5\\\implies x=3$

$\boxed{\boxed{\bold{Therefore, \ the \ required \ number \ is \ 35}}}}}}}$

                                                                                             

Answer 2

Answer:

35 is original number

Step-by-step explanation:

Let the digits in tens place be x and the digits in units place be y.

Therefore, the original two digit number is 10x + y.

Given that Sum of digits of a two digit number is 8.

= > x + y = 8 ------- (1)

Given that The new number increases by 18, when the digits are reversed.

= > 10y + x = 10x + y + 18

= > 10y - y = 10x - x + 18

= > 9y = 9x + 18

= > 9y - 9x = 18

= > y - x = 2

= > x - y = -2 ----- (2)

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On solving (1) & (2), we get

= > x + y = 8

= > x - y = -2

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2x = 6

x = 3.

Substitute x = 3 in (1), we get

= > x + y = 8

= > 3 + y = 8

= > y = 8 - 3

= > y = 5.

So,

= > 10x + y

= > 10(3) + 5

= > 35.

Therefore, the original number = 35.

Hope this helps!