Divisible by 3:
[125][/3] -> 41
Divisible by 7:
[125][/7] -> 17
Divisible by 21:
[125][/21] -> 5

Now the no.s divisible by both 3 and 7 are counted two times: in the list of no.s divisible by 3 and also in the list of no.s divisible by 7.
Therefore, we get 41 + 17 - 5 = 53

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The concept is straightforward; we just need to be super careful to ensure that we get the point.

From 1 to 125,
41 integers are divisible by 3.
17 integers are divisible by 7.
5 integers are divisible by 21.
Total is 41 + 17 - 5 = 53.

The answer is thus (B)

Tough one for me.

The basic formula is Total = A + B - (A&B) which translates into Total = (Divisible by 3) + (Divisible by 7) - (Divisible by 3 and 7)

A number that is divisible by 3 and 7 is asking for the LCM of 3 and 7. This is (3*7) = 21 since 3 and 7 are prime numbers an are thus prime factorized.

To find the values number of values divisible by 3, 7 and 21 we need to note that these are consecutive multiples. With that we can find the highest and lowest multiples from 1-125 and find our result.

3 : (123 -3)/3 + 1 = 41
7: (119-7)/7 + 1 = 17
21: (105-21)/21 +1= 5

Using the original equation, Total = 41+17-5 = 58 - 5= 53.

Asked: How many integers from 1 to 125, inclusive, are divisible by 3 or 7 or both?

Number divisible by 3 = {3,6,,,,,123} = (123-3)/3 + 1 = 41
Number divisible by 7 = {7,14,.....,119} = (119-7)/7 + 1 = 17
Number divisible by 3 & 7 or 21 = {21,....105} = (105-21)/21 + 1 = 5

Numbers divisible by 3 or 7 or both = 41 + 17 - 5 = 53

IMO B
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