Question

How many integers in the range 1 through 42 are divisible by 2, 3, or 7? anyone explain in detail by mutual exclusion closed

Comments

Yes this method will always valid, just use your set concept according to and/or.

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yes , so @joshi_nitish here or is mentioned so we have to simply sum it giving 41 ??   /

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@sumit,

please see solution given below by @Utkarsh and @Nikhil, they have given correct solution for your qsn.

Answer 1

let A be the set of numbers divided by 2,
B be the set of numbers divide by 3 and
C be the set of numbers divide by 7

A = {2,4,6,8 .......42 }

B = {3,6,9,....42 }

C = {7,14,21,28,35,42}

|A| = 21 , |B| = 14 , |C| = 6

By principle of inclusion and exclusion,

|A $\cup$ B $\cup$ C| = |A|+|B|+ |C| - |A$\cap$B| - |B$\cap$C| -  |A$\cap$C|  + |A$\cap$B $\cap$ C|

A$\cap$B = {6,12...42}
|A$\cap$B| = 7

B$\cap$C = {21,42}
|B$\cap$C| = 2

A$\cap$C = {14,28,42}
|A$\cap$C |  = 3

A∩B∩C = {42}
|A∩B∩C| =  1

 |A $\cup$ B $\cup$ C| = 21 + 14 + 6 - 7 - 2 - 3  + 1

 |A $\cup$ B $\cup$ C| = 30

Answer 2

Total number of integers = 42  (1 to 42)
Number of integers divisible by 2 = floor(42/2) = 21
Number of integers divisible by 3 = floor(42/3) = 14
Number of integers divisible by 7 = floor(42/7) = 6

Number of integers divisible by 2 and 3 both = floor( 42/ lcm(2,3) ) = 7
Number of integers divisible by 2 and 7 both = floor( 42/ lcm(2,7) ) = 3
Number of integers divisible by 3 and 7 both = floor( 42/ lcm(3,7) ) = 2

Number of integers divisible by 2, 3 and 7 all = floor( 42/ lcm(2,3,7) ) = 1
Now,
Number of integers divisible by either 2 or 3 or 7 = 21 + 14 + 6 - 7 - 3 - 2 + 1 = 30

Comments

42/3 = 14 .

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Thanks, I am correcting it.