The sum of a two digit number and the number obtained by reversing its digits is 121. The two digits differ by 3. Find the number.
47 or 74
Let the digits at the unit's place be x and the digit at the tens place be y. Then,
Number = 10y + x
From the condition given in the equation we have,
(10y + x) + (10x + y) = 121
⇒ 11(x+y)= 121 ⇒ x + y = 11
and, x - y = ± 3 [ Differences of digits is 3]
Thus, we have the following sets of simultaneous equations
x + y = 11 . . . (1) x - y = 3 . . . (2)
x + y = 11 . . . (3) x - y = -3 . . . (4)
On solving equations (1) and (2), we get
x + y = 11
x - y = 3
⇒ 2x = 11 + 3
⇒ x = 7
Substituiting x = 7 in equation (1)
7 + y = 11 ⇒ y = 4
∴ When x = 7, y = 4, we have
Number = 10y + x = 10 × 4 + 7 = 47
On solving equations (3) and (4), we get
x + y = 11
x - y = -3
⇒ 2x = 11 - 3
⇒ x = 4
Substituiting x = 4 in equation (3)
4 + y = 11 ⇒ y = 7
∴ When x = 4, y = 7, we have
Number = 10y + x = 10 × 7 + 4 = 74
Hence, the required number is either 47 or 74