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

Given, two-digit number is 4 more than 6 times the sum of its digits.

⇒ 10b + a = 6(a + b) + 4

⇒ 4b = 5a + 4 ------- (1)

Also, if 18 is subtracted from the number, the digits are reversed.

⇒ 10b + a – 18 = 10a + b

⇒ b – a = 2 ------- (2)

Multiplying eq2by 4 and subtracting from eq1

⇒ 4b – 5a – 4b + 4a = 4 – 8

⇒ a = 4

Thus, b = 6

Number is 64