Question

Find the least number which when divided by 4, 8, 12, 16, 20, 48 and 80 will leave in each case remainder 3.​

Answer 1

Answer:

I think Instead of 3, here it should be 2

Answer 2

The required number is 243.

Given:

Divisors = 4, 8, 12, 16, 20, 48, 80

Remainder = 3

To find: Dividend

Solution:

We know that

Least number perfectly divisible by n numbers = LCM of n numbers

⇒ Least number perfectly divisible by given numbers = LCM(given numbers)

⇒ Required number = LCM(given numbers) + 3

Now,

LCM(4, 8, 12, 16, 20, 48, 80) is needed.

List all prime factors for each number.

Prime Factorization of 4 is:2 x 2  =>  2²

Prime Factorization of 8 is:2 x 2 x 2  =>  2³

Prime Factorization of 12 is:2 x 2 x 3  =>  2² x 3

Prime Factorization of 16 is:2 x 2 x 2 x 2  =>  2⁴

Prime Factorization of 20 is:2 x 2 x 5  =>  2² x 5

Prime Factorization of 48 is:2 x 2 x 2 x 2 x 3  =>  2⁴ x 3

Prime Factorization of 80 is:2 x 2 x 2 x 2 x 5  =>  2⁴ x 5

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

⇒ The new superset list is 2, 2, 2, 2, 3, 5

Multiply these factors together to find the LCM.

⇒ LCM = 2 x 2 x 2 x 2 x 3 x 5 = 240

In exponential form: LCM = 24 x 31 x 51 = 240

⇒ LCM(4, 8, 12, 16, 20, 48, 80) = 240

⇒ Required number = 240 + 3 = 243

∴ The least number which when divided by 4, 8, 12, 16, 20, 48 and 80 will leave in each case remainder 3 is 243.