Q. 2

# Find the least nu

Let x be the required number.

Here, observe that the difference between the divisors and corresponding remainders is constant, that is –

20 – 14 = 25 – 19 = 35 – 29 = 40 – 34 = 6

As LCM is the smallest number that gives zero remainder when divided with all the numbers, the required number x is 6 less than the LCM of 20, 25, 35 and 40.

x = LCM of (20, 25, 35, 40) – 6

To find the LCM, each of the numbers is written as a product of the powers of its prime factors.

20 = 22 × 5

25 = 52

35 = 5 × 7

40 = 23 × 5

Now, LCM is the product of the highest powers of all factors.

Highest power of 2 = 3

Highest power of 5 = 2

Highest power of 7 = 1

So, LCM of (20, 25, 35, 40) = 23 × 52 × 7 = 1400

x = 1400 – 6 = 1394

Hence, the required number is 1394.

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