Let the one’s digit be ‘a’ and ten’s digit be ‘b’

Given, number is 3 more than 4 times the sum of its digits.

10b + a = 4(a + b) + 3

⇒ 10b + a = 4a + 4b + 3

⇒ a = 2b - 1 -----(1)

Also, if 18 is added to the number, the digits are reversed.

⇒ 10b + a + 18 = 10a + b

⇒ a – b = 2 ----- (2)

Subtracting eq2 from eq1

⇒ a – 2b – a + b = -1 – 2

⇒ b = 3

Thus, a = 6 – 1 = 5

Number is 35.