Question

A two digit number is 3 more than 4 times the sum of its digits if 18 is added to the number its digits are reversed find the number

Answer 1

Solution:
Let the digit in ones place be x and tens place be y
Original number = 10y + x
Number formed by reversing the digits = 10x + y
Given 10y + x = 4(x + y) + 3
⇒ 10y + x − 4x − 4y = 3
⇒ 6y − 3x = 3
⇒ 2y − x = 1 → (1)
Also given that when 18 is added to the number the digits gets interchanged.
Hence (10y + x) + 18 = 10x + y
⇒ 9x − 9y = 18
⇒ x − y = 2 → (2)
Add (1) and (2), we get
2y − x = 1
x − y = 2
---------------
y = 3
Put y=3 in x − y = 2, we get
x − 3 = 2
∴ x = 5
Hence (10y + x) = 10(3) + 5 = 35


OR

Let the digits be x and y.
Thus, number=10x+y
Now,
(10x+y) - 4(x+y)= 3.....(i)
and
(10x+y)+ 18=10y+x.....(ii)

Solving (i)
10x+y-4x-4y=3
or,6x-3y=3
or,2x-y=1....(1)
Solving (ii)
10x+y-10y-x=-18
or, 9y-9x=18
or,y-x=2...(2)

(1)+(2)
2x-y+y-x=1+2
or,x=3
Thus, y=x+2 (from (2))
y=5
Therefore the number= 10x+y=10x3+5=35