Math, asked by Allen27, 1 year ago

Please tell me the way to solve question no. 4 and 5

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Answered by Anonymous
0

This answer is of 4th

2520

We want LCM(1,2,3,4,5,6,7,8,9,10)

We can ignore 1, since any counting number is divisible by 1.

We prime factor each of the counting numbers from 2 to 10

2 = 2

3 = 3

4 = 2*2

5 = 5

6 = 2*3

7 = 7

8 = 2*2*2

9 = 3*3

10 = 2*5

The LCM of all those must have as many factors of

each prime that appears in any factorization

2 appears at most 3 times as a factor of 8

3 appears at most 2 times as a factor of 9

5 appears at most 1 time as a factor if 5 and 10

7 appears at most 1 time as a factor of 7

So the LCM has

3 factors of 2, 2 factors of 3, and 1 facor each of 5 and 7

LCM = 2*2*2*3*3*5*7 = 2520

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