Question

Find the smallest number which when divided by \[6,8,12,15\] and \[20\] leaves the same remainder 5.

Answer 1

Hint:
Least common multiple or smallest common multiple (denoted as L.C.M) of two integers (p and q) is the smallest integer that is divisible by both the integers p and q.
We can easily find the L.C.M of given integers simply by using the prime factorization method.

Complete step by step solution:
L.C.M of 6=\[6 = 2 \times 3\]
L.C.M of 8= \[2 \times 2 \times 2\]
L.C.M of 12= \[2 \times 2 \times 3\]
L.C.M of 15= \[5 \times 3\]
L.C.M of 20= \[2 \times 2 \times 5\]
L.C.M of 6, 8, 12, 15, 20 \[ = 2 \times 2 \times 2 \times 3 \times 5\]=120.
120 is the smallest number which will completely divide the numbers (6, 8, 12, 15, 20).
Since, we want remainder 5.

Therefore, the required smallest number is \[120 + 5 = 125\]

Note:
The below mentioned points will help you to solve this kind of questions:
1) Whenever you are required to find a number that is divisible by more than one number, find LCM.
2) Whenever you are required to find a number that completely divides more than one number, find HCF.