How many positive integers less than 5,000 are evenly divisi

Question here, why don't you multiply 47*2

I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this?
Thanks
Cheers
J

Hi Bunuel,
Is there a shorter way around this question?This approach took me 5 minutes!

Option B.(Took 5 minutes to solve with the following approach!)
All nos. from \(1\) to \(5000\) since we're given POSITIVE integers.\(=4999\)
Multiples of \(15\) in this range\(=4995/15=333\)
Multiples of \(21=4998/21=238\)
Multiples of both \(15\) and \(21\)=multiples of \(105=47\)
Now \(answer=4999-333-238+47=4475\)

This seems to be one of those killer problems that aren't necessarily conceptually difficult, but unless you have (what for me is) very high-powered arithmetic (long division) skills you're going to be in a time crunch here and in danger of compromising your quant section on test day. Practice, practice, practice those fundamentals!

One trick, because the ACs are so close together, is that you want to be careful to account for the fact that "less than 5000" accounts for only 4,999 integers, not including 5000. (Turns out on this problem it's not of such great concern, but on many other problems, it definitely is!) For this question, huge waste of about 20-30 seconds was the initial rephrase: "How many integers.. divisible by NEITHER 15 or 21." Don't know why but that tripped me up. Do they mean all of the numbers not divisible by 105 (Neither 15 AND 21, I saw the language trick there and ruled this out but not after wasting some time on math). What were are therefore looking for is Opposite: 4999 (minus) all the numbers that 15 or 21 go into (minus) overlap between multiples of 15 and 21. Let the long division begin!

hey Bunuel,
What does evenly divisible number means? I thought the quotient for such division should be multiple of 2.

Thanks,
Amit

Nope.

Evenly divisible means divisible without a remainder, so simply divisible.

In questions like these do we always have to take an LCM ?

I’m sorry for my ignorance, but why do we take the LCM of 15 and 21 to remove double count? I multiplied 15*21= 315 and divided 5000/315= 15 times to get the overlap. I know it is wrong now, but can someone explain it to me please?

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The numbers which are divisible by both 15 and 21 are numbers which are divisible by the least common multiple of 15 and 21, which is 105: 105, 210, 315, 420, ... If you simply multiply you are loosing many numbers, for example, 105, 210, 420, ...

We can find the no of integers divisible by 15 using the formula of A.P. with first term as 15 and common difference also 15.
Since, the largest term smaller than 5000 divisible by 15 is 4995.
4995 = 15 + (n-1)15. We get, n = 333.
Similarly, we can find no of terms divisible by 21.
So, no of integers from 0 to 5000 divisible by 21 = 238.
Now, we need to find the no of integers divisible by 15 x 21 because these integers are counted twice as the factors of 15 as well as the factors of 21.
No of factors of LCM of 15 &21 till 5000 = 47.
So, total no of positive integers till 5000 divisible by either 15 or 21 = 333 + 238 – 47 = 524.
Hence, no of integers not divisible by 15 or 21 = 4999 – 524 = 4475.

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