A number consists of two digits, whose sum is five. When the digits are reversed, the original number become greater by nine. Find the number.
A two-digit number has 2 digits, one at the unit's place and one at the tens place.
Answer: A two-digit number whose sum of the digits is five and when the digits are reversed, the original number becomes greater by 9 is 32.
The general form of a two-digit number with 'a' as the digit at the units place and 'b' as the digit at the tens place is:
10 × b + a = 10b + a
Let the digit at the unit's place be 'a' and the unit's at the ten's place be 'b', so the original number = 10b+a.
According to the conditions given in the question:
Sum of the two digits is five, that is:
a + b = 5 ---------------------- (1)
If the original number is 10b + a, numbers formed when the digits are reversed = 10a + b.
So the original number is greater than the reversed number by nine, that is:
(10b + a) - (10a + b) = 9
or, 10b + a - 10a - b = 9
or, 9b - 9a = 9
or, b - a = 1 ------------------ (2)
Now let us use the two equations (1) and (2) that we have got to find the values of a and b:
Form the first equation:
a + b = 5
or, a = 5 - b
Let's substitute this value of a in the second equation, as:
b - (5 - b) = 1
or, b - 5 + b = 1
or, 2b - 5 = 1
or, 2b = 1 + 5
or, 2b = 6
or, b = 3
Let's substitute this value of b in a = 5 - b to find the value of a:
a = 5 - 3 = 2
Substitute the values of a and b in the original number = 10b + a = 10 × 3 + 2 = 30 + 2 = 32.
Therefore, the original number is 32.