If 18 is added to a two-digit number, the digits are reversed. The sum of the digits is 8. What is the original number?

Question

Eli Assouline · Accepted Answer

#35#

Let #x# & #y# be respectively unit & tenth digits of two-digit number then the number will be

#10y+x#

Condition 1: sum of digits is #8# hence

#x+y=8\ .........(1)#

Condition 2: When #18# is added to the number #10y+x# then digits are reversed i.e. number becomes #10x+y# hence

#18+10y+x=10x+y#

#9x-9y=18#

#x-y=2\ ..........(2)#

By adding (1) and (2) in this manner

#x+y+x-y=8+2#

#2x=10#

#x=5#

Setting #x=5# in (1), we get

#5+y=8#

#y=3#

Consequently, the figure is

#10y+x#

#=10\cdot 3+5#

#=35#

Eva Abramson · Answer

undefined Let the original two-digit number be represented as 10x + y, where x is the tens digit and y is the units digit. According to the given conditions, the reversed number is represented as 10y + x.

So, the equation becomes:
10x + y + 18 = 10y + x

Simplifying the equation:
9x - 9y = 18
x - y = 2

Since the sum of the digits is 8, we have:
x + y = 8

Now, we can solve the system of equations:
x - y = 2
x + y = 8

Adding the two equations:
2x = 10

Dividing both sides by 2:
x = 5

Substituting the value of x into one of the equations:
5 + y = 8
y = 3

So, the original number is 53.