Q. 2

Find the least nu

Answer :

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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