Question

The smallest three digit number when divided by 12, 16 and 18 leaves in each case a remainder 5, is
$\left( a \right)149 \\\ \left( b \right)293 \\\ \left( c \right)337 \\\ \left( d \right)481 \\\$

Answer 1

Hint: In this question, we have to find the L.C.M. of 12, 16 and 18 and then add reminder (5) in the LCM. Least Common Multiple (LCM) is a method to find the smallest common multiple between any two or more numbers.

Complete step-by-step answer:

Given, we have three numbers 12, 16 and 18.
To find three digit numbers we have to find the L.C.M. of 12, 16 and 18 and then add reminder (5) in the LCM.
Required three digit number = (LCM of 12, 16 and 18) + Remainder
Now, we have to find LCMs of 12, 16 and 18.
The factor of $12 = 2 \times 2 \times 3$
The factor of $16 = 2 \times 2 \times 2 \times 2$
The factor of $18 = 2 \times 3 \times 3$
LCM is the smallest common multiple between any two or more numbers.
So, the LCM of 12, 16 and 18 $= 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 144$
Now, we have to find the smallest three digit number when divided by 12, 16 and 18 leaves in each case a remainder 5.
So, Required three digit number = (LCM of 12, 16 and 18) + Remainder
Required three digit number=144+5=149
So, the correct option is (a).

Note: Whenever we face such types of problems we use some important points. Like first we find the factors of all given numbers then find the LCM of given number with help of factors. So, after adding the remainder in the LCM we will get the required answer.