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There was an event held by a group. In this COVID-19 era, there were many hardships in equipping quarantine supplies, in addition to carrying out the event. Several bottles of the same hand sanitizer were purchased for those attending the event. One bottle of the sanitizer for every ten people was equipped. One to ten people need a bottle of the sanitizer, but if they are eleven, two bottles are needed.

Initially, hand sanitizers were prepared according to the number of people scheduled to attend, but on the day of the event, exactly one-third of the scheduled participants were absent, so 12 bottles of the sanitizer did not need to be taken out of the warehouse.

At most, how many people were expected to attend this event?

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    $\begingroup$ What part of 'global pandemic' do these people not get? $\endgroup$
    – Strawberry
    Feb 9 at 12:25
  • $\begingroup$ More than the legally prescribed maximum for a public gathering. $\endgroup$ Feb 9 at 19:48
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    $\begingroup$ @user3067860 Legally prescribed in which country? $\endgroup$ Feb 10 at 8:32
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Non-programming solution

Say there were originally $3N$ people scheduled to attend the event, but only $2N$ actually came. Then

the number of bottles originally bought was $\lceil\frac{3N}{10}\rceil$ and the number actually needed was $\lceil\frac{2N}{10}\rceil$,

so we must solve the equation

$$\lceil\frac{3N}{10}\rceil-\lceil\frac{2N}{10}\rceil=12.$$

Now we have

$\frac{2N}{10}+1\;{\color{red}\geq}\;\lceil\frac{2N}{10}\rceil=\lceil\frac{3N}{10}\rceil-12\;{\color{blue}\geq}\;\frac{3N}{10}-12$, so $2N+10\geq3N-120$, i.e. $N\leq130$.

This gives us an upper bound.

If $N=130$ (a multiple of 10), then the blue inequality is achieved as equality but the red one isn't, but the overall inequality is - contradiction.
So $N$ must be sufficiently far below $130$ that either $\lceil\frac{3N}{10}\rceil$ or $\lceil\frac{3N}{10}\rceil$ is different from what it was when $N=130$. That means $N$ must have moved to the other side of a multiple of either $\frac{10}{3}$ or $\frac{10}{2}$, so the highest possibility for $N$ is $N=126$.

Now a quick calculation (just one calculation!) verifies that this is indeed the maximal possible solution, which means the answer is

$3\times126=378$.

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  • $\begingroup$ And at the least, it is 342, right ? $\endgroup$ Sep 2 at 12:08
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If I understand correctly, then we are looking at solutions to the following equation:

$$\lceil \frac{3N} {10} \rceil - \lceil \frac {2N} {10} \rceil = 12.$$

It is easy to get all possible values of $N$:

$$N = 114, 115, 117, 118, 119, 120, 121, 122, 123, 126.$$

This step could be done by hand, e.g. by considering different possibilities of $N \mod 10$. But I just wrote a quick program, as it's more efficient.

Thus at most there are

$$3 \times 126 = 378$$

expected participants.

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Mental math solution:

There were 3 * 126 = 378 people expected at the party.

To get there with fairly easy mental math:

Focus on the number of people who didn't show up. 12 unneeded bottles means around 12 * 10 = 120 people were no-shows. If exactly 120 people didn't show up, then there would have been 12 fewer bottles needed no matter the exact number of people who did show up. Similarly, if 130 people didn't show up, there would have been 13 fewer bottles needed. So the maximum number of no-shows must be at least 120 and less than 130.

From there, we can simply enumerate a few possibilities:

We'll say there were 120 + N no-shows (and therefore 360 + 3N expected attendees) so that we can focus on the ones digit. If N is greater than the ones digit of 3N, then it's going to cause the total to roll over and result in 13 unneeded bottles. So we can quickly say that N = 9, 3N = 27; N = 8, 3N = 24; and N = 7, 3N = 21 all won't work. Then we come to N = 6, 3N = 18 and see that it will work, so we have 120 + 6 no-shows of the 378 originally expected attendees.

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An alternative solution with mostly math and minimal case analysis. (I initially aimed for "minimal textual explanation", but apparently I failed)

As other answers have already pointed out, we can start with the equation

$$ \left\lceil \frac{3N}{10} \right\rceil - \left\lceil \frac{2N}{10} \right\rceil = 12 $$

Now, let

$N=5q+r$, where $0 \le r < 5$.

Then we get

$$ \left\lceil \frac{3(5q+r)}{10} \right\rceil - \left\lceil \frac{2(5q+r)}{10} \right\rceil = 12 \\ \left\lceil \frac{3}{2}q+\frac{3}{10}r \right\rceil - \left\lceil q+\frac{r}{5} \right\rceil = 12 $$

Since we can factor out integral parts (namely $q$) from the ceilings,

$$ \left\lceil \frac{q}{2}+\frac{3}{10}r \right\rceil - \left\lceil \frac{r}{5} \right\rceil = 12 $$

The maximal possible value of $q$ is

$q=25$, because $\frac{3}{10}r \ge \frac{r}{5}$, so if $\frac{q}{2} \ge 13$ then the LHS would be always at least 13.

If we plug it in, we get

$$ \left\lceil \frac{1}{2}+\frac{3}{10}r \right\rceil - \left\lceil \frac{r}{5} \right\rceil = 0 $$

Then

the first term is at least 1 (because of $\frac{1}{2}$) and the second is at most 1 (because $0 \le r < 5$), so the only possible value for both terms is 1.

Finally we solve

$$ \frac{1}{2} + \frac{3}{10}r \le 1 \Rightarrow r \le \frac{5}{3} \\ \frac{r}{5} > 0 \Rightarrow r > 0 $$

which gives

that the only possible integral value of $r$ is $r=1$, and therefore $N=126$. The expected participants were $3N = 378$.

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  • $\begingroup$ @Downvoter, can you elaborate? $\endgroup$
    – Bubbler
    Feb 10 at 5:49
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The original number can be written as something like $30k+3r$ ($r<10$). Then the remaining number of people must be $20k+2r$.

So, there are $3k+\left\lceil3r/10\right\rceil$ sanitizers among which only $2k+\left\lceil2r/10\right\rceil$ had to be used.

$3k+\left\lceil3r/10\right\rceil$ - $(2k+\left\lceil2r/10\right\rceil) = 12$

$k+\left\lceil3r/10\right\rceil - \left\lceil2r/10\right\rceil = 12$

Since $\left\lceil3r/10\right\rceil >= \left\lceil2r/10\right\rceil$, the best case is $k=12$ and $\left\lceil3r/10\right\rceil = \left\lceil2r/10\right\rceil$.

If $2r = 10x+2y$ and $3r = 15x+3y$ with $y<5$, then:

$15x+3y = 10x+2y + (5x+y)$

$5x+y<10$, $10x+2y<20$, $15x+3y<30$

$\left\lceil2r/10\right\rceil<3$, $\left\lceil3r/10\right\rceil<3$, $15x+3y<=20$

$15x+3y=18$ can satisfy these conditions. So, with $r=6$ and $k=12$, we get $378$.

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Unlike others, my solution is

240

Reason:

While 12 bottles are left over, and these are exactly 1/3 of all bottles needed. That means that (12x3)-(12x1) = 24 bottles of sanitizer would be used for the party. As 10 People is the most we can assume for a bottle, we have 24x10 = 240

If we would have 241 Participants, a new bottle would be needed, so from 36 bottles, we would subtract 25 for the party, which means that only 11 bottles would be left in storage

PS:

The minimum participants are 231, but that wasn't asked

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  • $\begingroup$ But what if 242 actually attended, then there would be 363 originally scheduled to attend, and you would use 25 bottles instead of the planned 37, which is still a saving of 12 bottles. Your assumption that 12 is exactly one third of the planned number of bottles is not correct. $\endgroup$ Feb 10 at 12:53
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My answer for the number of attendees is minimum 232 maximum 240.

People attending: minimum 232, maximum 240

Absentees: minimum 116, maximum 120

NOTE: 231/2=115.5 and 230/10=23

In my scenario, the number of bottles on hand is 36, because 24 bottles were opened when the number of attendees is minimum 232 and maximum 240. So 24 + 12 =36. Another scenario is to take the number of absentees minimum 120 maximum 128, with the attendees are minimum 240 and maximum 256. If we take the maximum, 26 bottles have to be opened, and 26 + 12 =38. bottles on hand. In the first case the number of bottles on hand is always 36. This is not true for the second scenario.

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    $\begingroup$ Could you explain how you got these numbers? $\endgroup$
    – bobble
    Feb 10 at 3:56
  • $\begingroup$ For every 10 people, 1 bottle of sanitizer is needed. There were 12 unused bottles, which means the number of absentees is between 116 and 120. The NOTE explicitly gives the numbers justifying the minimum. $\endgroup$ Feb 10 at 4:34
  • $\begingroup$ @VassilisParassidis That's not necessarily true. If there were 121 scheduled participants (requiring 13 bottles) but only 10 showed up (requiring 1 bottle), there were only 111 absentees. Similarly with 130 scheduled and 1 showing up, you get 129 absentees. $\endgroup$
    – Bubbler
    Feb 10 at 5:01
  • $\begingroup$ 1/3 of the scheduled participants were absent so 12 bottles were not needed. Since you think the total number of participants was 121, that should have been in your answer. $\endgroup$ Feb 10 at 7:38
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    $\begingroup$ @VassilisParassidis No, I don't mean 121 is the answer. I was just giving examples where you save 12 bottles but the number of absentees is not between 116 and 120. My answer of 378 (which all the other answers also agree) meets all the requirements: 378 requires 38 bottles, but if 1/3 of that is absent, we have 252 participants which require 26 bottles, so we save 12 bottles. $\endgroup$
    – Bubbler
    Feb 10 at 8:16

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