A boarding school accommodating 400 students has provided a locker to

Total students: \(400\)

Single gift
Total students who got Pens: \(\frac{400}{2}=200\)
Total students who got Chocolates: \(\frac{400}{3} = 133.33\) round down to make it \(133\)
Total students who got Wine: \(\frac{400}{5} = 80\)

Two gifts
Total Students who got Pens & Chocolates: \(\frac{400}{6}= 66\)
Total Students who got Pens & Wine: \(\frac{400}{10}= 40\)
Total Students who got Wine & Chocolates: \(\frac{400}{15}= 26\)

All three gifts
Total Students who got Pens, Chocolates & Wine:\(\frac{400}{30}= 13\)


Working out how many students received no gifts
Total Students = (Sum of Single Gifts) - (Sum of Two Gifts) + (Sum of three gifts) + (no gifts)

\(400 = (200 + 133 + 80) - (66 + 40 + 26) + 13 + n\)

\(400 = 413 - 132 + 13 + n \)

\(400 = 294 + n\)

\(n =106\)

Answer C

Why do we add the "sum of three gifts" and not subtract it as we do with the "sum of two gifts"?

I am missing something here. I am getting 145 as the answer. (I am subtracting 2*13 as all three). Can someone help me what piece am I missing ?

kajumba900 prachisaraf

Yeah, this is a pretty tricky variation on a standard 3-set question. Specifically, each of the groups Nidzo identifies contains the following ones. In other words, the count of people who got one gift includes all the people who got 2, which in turn includes all the people who got 3.

If we knew the number of people who got EXACTLY two or EXACTLY 3, we could use a more standard formula:

Total = group 1 + group 2 + group 3 - doubles - 2(triples)

The idea there is that if I am in 2 groups, I got counted twice, so we subtract the number of people in 2 groups. If I am in 3 groups, I got counted 3 times, so we subtract TWICE the number of people in 3 groups. In this case, however, since the people in 2 groups were *also* in 3 groups, we subtracted them too many times. For instance, imagine I got all 3 gifts. I got counted 3 times in the first count (got pens, got pencils, got chocolates). Then I got SUBTRACTED 3 times in the second count (pens+choc, pens+wine, wine+choc). So now it's like I never got counted at all, and you have to add me back in! Alternatively, you could say that I got subtracted three times when I should only have been subtracted twice.

Let me know if that made sense. I'm still trying to understand why Santa is giving wine to kids and then leaving ~1/4 of the kids empty-handed. Does he have a grudge against prime-numbered lockers?
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I also thought of an interesting "brute force" way to work this out. We've seen that the gifts all line up at every 30th locker, right? Since 30 = 2*3*5, every time we get to a multiple of 30, the pattern will basically reset as if we're counting from 1. So we can look at the first 30 lockers and extrapolate.

Start with just the odds:

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29

Now cut the 3's and 5's:

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29

That leaves us with 8 empty lockers: 1, 7, 11, 13, 17, 19, 23, 29

We'll have this same thing in every group of 30. (Don't believe me? If we add 30 to all our numbers in the first set, that doesn't change whether they are multiples of 2,3, or 5. Try it!)
If we look at 13 groups of 30 lockers, that will take us almost to the end: locker 390.

So far we have 13 groups of 8 empty lockers. 8*13 = 104 empty lockers. At this point, only C makes sense. There are just 10 lockers left, and fewer than half of them will be empty, so we can't get to 110. If we wanted to be sure we were on the right track, we could write them out:

391 393 395 397 399

395 ends in 5, so that's out. 390 was a multiple of 3, so the next ones are 393, 396, 399. What's left?

391 393 395 397 399

Just two lockers: 391, 397.

So our grand total of empty lockers is 104+2 = 106. There we go!
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