In a 3-digit number, the ten's digit is thrice the unit digit and hundred's digit is twice the unit digit. If the sum of all the three digits is 12, find the number.

Let the 3-digit number be abc.

abc = 100a + 10b + c

Given,

b = 3c ……..(1)

a = 2c ……..(2)

and

a + b + c = 12 ……..(3)

Putting value of b and a from eqns (1) and (2) into eq (3) we get,

2c + 3c + c = 12

⇒ 6c = 12

⇒ c = $\dfrac{12}{6}$

⇒ c = 2

Putting value of c in eq 1,

b = 3c = 3 x 2 = 6

Putting value of c in eq 2,

a = 2c = 2 x 2 = 4

Hence, the number is 462.