Question

Find the least number which when divided by 40, 48 and 45 leaves reminder 3 in each case.

Answer 1

LCM of 40 = 2 * 2 * 2 * 5

LCM of 48 = 2 * 2 * 2 * 2 * 3

LCM of 45 = 3 * 3 * 5

LCM of 40,48,45 = 2 * 2 * 2 * 2 * 3 * 3 * 5

                            = 720.

Add remainder to the LCM.

I.e 720 + 3 = 723.

Answer 2

The least number which when divided by 40, 48 and 45 leaves reminder 3 in each case is 723.

Given:

40, 48 and 45

To find:

Find the least number which when divided by 40, 48 and 45 leaves reminder 3 in each case.

Solution:

To fiend the least number first we have to find the LCM of 40, 48 and 45

First we will find prime factors of these numbers

40 = 2*2*2*5 = 2³*5

45 = 3*3*5 = 3² *  5

48 = 2*2*2*2*3 = $2^{4}$ * 3

Now, we will take highest powers of 2,3 and 5 which are $2^{4}$,  3² and 5 and now we will multiply them

LCM = $2^{4}$ * 3² *5 = 2*2*2*2*3*3*5 = 720

Now, we will add remainder to the LCM

Final answer = 720 + 3 = 723

Hence, the answer is 723.

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