Question

What is the smallest number, which when divided by 7, 18, 56 and 36, leaves a remainder zero?

Answer 1

Answer:

Step-by-step explanation:According to me, the smallest number should be the LCM of all these numbers.

Therefore, 3*2*3*7*2*2=504

Answer 2

Given: Numbers - 7, 18, 56 and 36

To find: The smallest number which when divided by given numbers leaves remainder 0.

Solution: The smallest number which is exactly divisible by 7, 18, 56 and 36 is their Least Common Multiple.

Finding LCM of 7, 18, 56 and 36 using Prime factorization Method.

7 = 7 × 1

18 = 2 × 3 × 3 × 1

56 = 2 × 2 × 2 × 7 × 1

36 = 2 × 2 × 3 × 3 × 1

LCM is expressed as the product of maximum frequencies of all the prime factors.

Now, 7 occurs maximum 1 time, 2 occurs maximum 3 times and 3 occurs maximum 2 times.

LCM = 2 × 2 × 2 × 3 × 3 × 7 = 504

Hence, the smallest number which is exactly divisible (leaves remainder 0) by 7, 18, 56 and 36 is 504.