How many positive integers less than 500 are divisible by 6, 8 or 10?

The question can be solved using venn diagram approach. The only catch in the question is that the factors are not prime. So while subtracting the overlaping part we have to take the LCM unlike multiplication in prime numbers. Because 24 will also be divisible by both 6 and 8 it need not necessarily be 48.

So what we need to see is ==>

Numbers divisible by 6
+ Numbers divisible by 8
+ Numbers divisible by 10
- Numbers divisible by 24
- Numbers divisible by 40
- Numbers divisible by 30
+Numbers divisible by 120

Which is

50+62+83-16-12-20+4
=151



Also for this solution question should be read as "How many positive integers less than or equal to 500 "

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Number of integers divisible by 6: 83
Number of integers divisible by 8: 62
Number of integers divisible by 10: 50
Total: 195
Take out the duplicates in combination of {6,8}.{8,10},{6,10},{6,8,10}: 24
Total distinct positive integers: 171.

Bunuel VeritasKarishma please correct me here if I am wrong.

How did you take out the duplicates?

A number divisible by both 6 and 8 which would be double counted would be divisible by 24, the LCM of 6 and 8. So you need to remove all multiples of 24. Similarly for 8 and 10, you need to find the multiples of 40 and for 6 and 10, you need to find the multiples of 30.
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Yes because otherwise I was working with 1-499, which is not wrong at all but then we need 150 in the answer choice.
The only difference is when 10 is taken on its own, as shown -
when divided by 6 --> 83
when divided by 8 --> 62
when divided by 10 --> 49
when divided by lcm(6,8) --> 20
when divided by lcm(8,10) --> 12
when divided by lcm(6,10) --> 16
when divided by lcm(6,8,10) --> 4

Answer = 83+62+49-20-12-16+4=150

I also took the exact same venn diagram approach.

Let m(a), m(a & b), m(a or b) denote the number of multiples of a, a and b, and a or b, respectively, that are less than 500. We can use the formula:

m(6 or 8 or 10) = m(6) + m(8) + m(10) - m(6 & 8) - m(6 & 10) - m(8 & 10) + m(6 & 8 & 10)

We have:

m(6) = (498 - 6) / 6 + 1 = 83


m(8) = (496 - 8) / 8 + 1 = 62


m(10) = (490 - 10) / 10 + 1 = 49

m(6 & 8) = m(24) = (480 - 24)/24 + 1 = 20

m(6 & 10) = m(30) = (480 - 30)/30 + 1 = 16

m(8 & 10) = m(40) = (480 - 40)/40 + 1 = 12

m(6 & 8 & 10) = m(120) = (480 - 120) / 120 + 1 = 4

Therefore,

m(6 or 8 or 10) = 83 + 62 + 49 - 20 - 16 - 12 + 4 = 194 - 48 + 4 = 150

Answer: 150

(Note: the official answer is D: 151. However, the problem should be stated as “less than or equal to 500” instead of “less than 500.” If that is the case, m(10) = 50, instead of 49.)
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There is no way this can be solved under 2 mins

Actually it is not very hard because the numbers are from 1 to 499 (these are 499 numbers).
(Something like 12 to 510 would have given more heartache.)

With the range of 1 to 499, if you need to find multiples of 6, you just divide 499 by 6, you get 83.(something). You can happily ignore the something and 83 multiples is your answer. This is what you do for all divisors and you get your answer.

Now consider the case with the range that doesn't start with 1. Say 12 to 510 (again 499 numbers)
But here, you can't simply say 499/6 = 83 multiples
6*2 = 12
6*85 = 510
So you have total (85 - 2) + 1 = 84 multiples of 6 in this range. This is what you would need to do for all divisors!
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VeritasKarishma Does this sort of problem appear in the real gmat test? Thanks for the concepts to perform the problem promptly.

Yes, this problem uses concepts routinely tested in GMAT. Though, the framing of the question may be more interesting while requiring fewer calculations.
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­Since in the question it is mentioned that all the positive integers less than 500, that means we cannot include 500 as a multiple of 10 for calculation purposes. In that case the answer I am getting is 150 and not 151. Is there anything that I am missing?
 

­
Yes, couple of posts above mention this too.­ Fixed the issue in the stem.
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Why are we adding the numbers divisible by 120? Why are we not subtracting it likewise in the case of LCM(8,6) = 24 etc?

­Each of the first six terms accounts for numbers divisible by 120—specifically, 120, 240, 360, and 480. When you sum the counts from the first three groups and then subtract the counts from the next three, you effectively exclude all multiples of 120. To correct this and ensure all relevant numbers (120, 240, 360, and 480) are accurately accounted for, you add those four numbers back in.

Hope it's clear.
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n(A U B U C)=n(A)+.. - n(A&B) +n(A&B&C)
= [500/6]+[500/8]+[500/10]-[500/LCM(8,6)]-[500/lcm(6,10)]-[500/lcm(8,10)]+[500/lcm(8,6,10)]
=195-48+4=151

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