The sum of the digits of a certain two-digit number is 9. When you reverse its digits you increase the number by 63. What is the number?

The given such number is 18.

Step-by-step explanation:

Here, sum of the two digits is 9

Let us assume the tens digit = m

So, the unit's digit of the number = 9 - m

Now, the current value of the number = 10 (Tens Digit) + 1 (units digit)

= 10 (m) + 1 (9-m) = 10 m + 9 - m = 9 m + 9 ... (a)

Second Case:

Here after reversing the digits, the ten's digit = 9 - m

The unit's digit = m

So, the new value of the number = 10 (Tens digit) + 1 (Unit Digit)

= 10 (9 - m) + m = 90 - 10 m + m = 90 - 9 m ... (b)

Now, according to the question:

The initial Number value + 63 = The new number value

or, 9 m + 9 + 63 = 90 - 9 m

or, 18 m = 99 - 72 = 18

or, m = 18/18 = 1

or, m = 1

Hence the tens digit of the number = m = 1

Also, the ones digit = 9 - 1 = 9 - 1 = 8

So, the given number is 18