A boarding school accommodating 400 students has provided a locker to

Question

A boarding school accommodating 400 students has provided a locker to each student. On Christmas eve, Santa places a pen in every other locker, a box of chocolate in every third locker and a bottle of wine in every fifth locker starting with the second locker for a pen, the third locker for a box of chocolate and fifth one for a bottle of wine. How many students have got none of the three gifts?

A. 120
B. 110
C. 106
D. 90
E. 56

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Official Answer

C

A. 120
B. 110
C. 106
D. 90
E. 56

Are You Up For the Challenge: 700 Level Questions

Nidzo · Accepted Answer

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

kajumba900 · Answer

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

prachisaraf · Answer

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 ?

DmitryFarber · Answer

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?

DmitryFarber · Answer

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!

HarshavardhanR · Answer

Lockers receiving pens -> 2, 4, 6, ...400 -> 200 lockers 
 Lockers receiving chocolates -> 3, 6, 9, ....399 -> 133 lockers 
 Lockers receiving wine -> 5, 10, 15, .....400 -> 80 lockers

We need to find the number of lockers that did not receive any gift.

Initially, we have (1, 2, ...400) -> 400 
 All even number lockers receive gifts. Exclude these. Remaining lockers -  (1, 3, 5, ... 399) -> 200  (all odd numbers)

All 3 multiples among these odd numbers receive gifts. Exclude these.  (3, 9, ...399) -> 67  numbers to be excluded. 
 So, still remaining ->  200 - 67 = 133 numbers

All 5 multiples receive gifts too. They should be excluded.  (5, 15, 25, ....395) -> 40  numbers to be excluded 
 If we remove these numbers, we will get  133 - 40 = 93 numbers

But this is where we need to exercise caution . When we removed 3-multiples, we also removed those 3-multiples which are also 5-multiples. These are (15, 45, 75, ....375). So, when we now removed the 5 multiples (by doing 133-40), these numbers got removed a second time. Not ideal!

To adjust for this, we have to add back these numbers one time.

So, to remove the double counting of (15, 45, ...375), add back these numbers -> 13 numbers 
 So, remaining numbers ->  93 + 13 = 106 numbers

Now, we have removed all 2-multiples, 3-multiples, and 5-multiples, without any double/triple counting. Our answer is  106. Choice C .

---
Harsha

DyutiB · Answer

As mentioned in the question, every 2 nd  lockers starting from 2 receive a pen, i.e the lockers receive a pen are – 2,4,6,........,400 (Arithmetic progression with difference 2). So the total number of lockers receive pens = 200. Also the lockers are multiple of 2 (an even number)
So a+d+e+g = 200
Similarly, every 3 rd  lockers starting from 3 receive a chocolate, i.e the lockers receive a chocolate are – 3,6,9, ........,399 (Arithmetic progression with difference 3). So the total number of lockers receive chocolate = 133. Also the lockers are multiple of 3.
So b+e+f+g = 133
Again, every 5 th  lockers starting from 5 receive a wine bottle, i.e the lockers receive a wine bottle are – 5,10,15,........,400 (Arithmetic progression with difference 5). So the total number of lockers receive wine bottles = 80. Also the lockers are multiple of 5.
So, c+d+f+g = 80
Now the common multiple of 2 and 3 is 6. So all the multiples of 6 will have both pen and chocolate. So, the lockers that have both pen and chocolate are – 6,12,18,........,396. (Arithmetic progression with difference 6). So the total number of lockers receive both pen and chocolate = 66. 
So, e+g = 66
Similarly the common multiple of 3 and 5 is 15. So all the multiples of 15 will have both chocolate and wine. So, the lockers that have both chocolate and wine are – 15,30,45,........,390. (Arithmetic progression with difference 15). So the total number of lockers receive both pen and chocolate = 26. 
So, f+g = 26
Now the common multiple of 2 and 5 is 10. So all the multiples of 10 will have both pen and wine. So, the lockers that have both pen and wine are – 10,20,30,........,400. (Arithmetic progression with difference 10). So the total number of lockers receive both pen and wine = 40. 
Now the common multiple of 2, 3 and 5 is 30. So all the multiples of 30 will have all of three- pen, chocolate, and wine. So, the lockers that have all three are – 30,60,90,........,390. (Arithmetic progression with difference 6). So the total number of lockers receive both pen and chocolate = 13. 
So, g = 13. 
Which gives us, e = 53, f = 13, d = 27.
And as the questions says, a+b+c+d+e+f+g+h = 400.
Adding up, a+b+c+2(d+e+f)+3g = 200+133+80 = 413.
i.e. 400+d+e+f+2g = 413+h
i.e. 400+27+53+13+26 = 413+h
i.e. h = 106 
Ans: 106 (C)

XLmafia21 · Answer

Thank you. this is a pretty solid and logical approach to not kill time

A boarding school accommodating 400 students has provided a locker to

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