Question

The sum of the digits of a two-digit number is 10. When the digits are reversed, the new number is 18 less than the original number. Find the original number. Check your answer. PLEASE HELP!!

Answer 1

64 is the original
46 is the reversal

64 -46 = 18
6 + 4 = 10

Answer 2

ANSWER:
The original number is 64

SOLUTION:
Let the two-digit number be ab, where 'a' is the tens digit and 'b' is the units digit.
"The sum of the digits of a two-digit number is 10" implies
a + b = 10 (Equation 1)

"When the digits are reversed, the new number is 18 less than the original number" is algebraically written as
ba = ab - 18
>> bx10 + ax1 = (ax10 + bx1) - 18
>> 10b + a = 10a + b - 18
>> 9a - 9b = 18
>> 9(a - b) = 18
>> a - b = 2
>> a = b + 2 (Equation 2)

Now substitute equation 2 in equation 1 to get
a + b = 10
>> (b + 2) + b = 10
>> 2b = 8
>> b = 4

Plug this value back in equation 1 to get
a = b + 2 = 4 + 2 = 6

So a = 6 and b = 4. Hence, the original number is 64.

CHECK:
Given: The sum of the digits of a two-digit number is 10
Check: 6 + 4 = 10 (so it is true)

Given: When the digits are reversed, the new number is 18 less than the original number
Check:
Digits reversed: 64 reversed becomes new number 46, and 46 is 18 less than 64, that is, 46 = 64 - 18.