Question

Find the smallest number which, when increased by 3 is exactly divisible by 15, 21, 45, 24 and 36

Answer 1

Answer:

2523

Step-by-step explanation:

prime factors of 15,21,45,24 and 36

15=3*5

21=3*7

45=3*3*5

24=2*2*2*3

36=2*2*3*3

LCM=2*2*2*3*3*5*7

=8*9*5*7=2520

Thus th eno=lcm+3=2523

Answer 2

Answer:

2517

Solution:

LCM of 15, 21, 45, 24 and 36 is 2520.

The value which is 3 less than the lcm of 15, 21, 45, 24 and 36 is

= 2520 - 3

= 2517

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LCM - lowest common multiple, lcm of two( or more) numbers is the smallest number which is evenly divisible by both the numbers.

Suppose we have to find LCM of a and b

To find LCM, prime factorise two( or more) numbers)

a = c × f × g

b = c × j × m

Find the product of multiples of both the numbers taking common multiple once.

LCM of a and b = c × f × g × j × m