@Rohan Rao
–
They didn't even bother to change the number of people it's been solved by..I don't like remembering problems I like solving them..Please brilliant,If you are going to recycle problems,At least change the numbers

They are probably behind, or saving something because of the revamp of curriculum mathematics that is being done. So I bet they are working very hard this week!
Lets find problems other places. If you have any good problems you would like to share, please post them as a reply so that everyone can have problems to do while waiting.

Here,
An auditorium has a rectangular array of chairs. There are exactly 14 boys seated
in each row and exactly 10 girls seated in each column. If exactly 3 chairs are
empty, find the maximum number of chairs in the auditorium.

Under the assumption that boys and girls cannot share a chair, let number of rows and columns be r and c. (r >= 14; c >= 10)

Then rc = 3+14r+10c

rc-14r-10c-3 = 0

rc - 14r - 10c +140 - 143 = 0

(r-10)(c-14) = 143

Now, the possibilities for (r,c) are (1,143), (11,13), (13,11) and (143,1). Checking all 4, the number of chairs in each is (11)(157), (21)(27), (23,25), (153,15).

Clearly (23)(25) > (21)(27) and (11)(157)<(153)(15). (if a>c>d>b>0 and a+b = c+d, ab < cd)
Also, (153)(15) > (23)(25) obviously.

The maximum number of chairs is thus (153)(15) = 2295 (with 2292 chairs filled up)

why ? defining X = r-10, Y = c-14, then that means
X = 1,Y = 143 -> r = 11, c 157 -> rc = 1727
X = 11, Y = 13 -> r = 21, c = 27 -> rc = 567
X = 13, Y = 11 -> r = 23, c = 25 -> rc = 575
X = 143, Y = 1 -> r = 153, Y = 15 -> rc = 2295 (This is the max number of chairs)

@Raymond Christopher Sitorus
–
I think the answer would be rc and not rc-3 though, since the three chairs that are empty are still chairs in the array.

There will be a lot to look forward to; new challenges and a huge database will be up tomorrow or Tuesday. Peter T. posted this earlier. I can't find a link at the moment, sorry!

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestOh man..I even woke up early to do the new problems today..

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same over here.. sad. :(

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Hi everyone,

This was a glitch and has been corrected. You can see new Olympiad problems now. We are very sorry for the mistake.

Happy problem solving!

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why are some of them recycled though??Ive seen them before..or is it just me

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Yes, i too saw at least 2 repeated questions.

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Yeah...

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They are probably behind, or saving something because of the revamp of curriculum mathematics that is being done. So I bet they are working very hard this week! Lets find problems other places. If you have any good problems you would like to share, please post them as a reply so that everyone can have problems to do while waiting.

Here, An auditorium has a rectangular array of chairs. There are exactly 14 boys seated in each row and exactly 10 girls seated in each column. If exactly 3 chairs are empty, find the maximum number of chairs in the auditorium.

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If you wanna keep this going, then here's a GREAT problem from the 1983 ARML competition:

In an isosceles triangle, the altitudes intersect on the inscribed circle. Compute the cosine of the vertex angle.

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Should I post a solution for this one? Its a fairly simple exercise in trigonometry, but there will be people still trying to solve this one.

I arrived at the answer 1/9 (assuming you are taking the cosine of the 'unique' angle)

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Don't post the solution though; while its a relatively simple problem I still like it for having such an interesting condition.

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Thanks for posting! :)

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try this: 1=6 2=12 3=18 4=24 5=30 6=??

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1,because 1=6,so 6=1 :)

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Yes 6=6 obviously but if the equal to mean's multiply by 6 its 36

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1 is correct

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Under the assumption that boys and girls cannot share a chair, let number of rows and columns be r and c. (r >= 14; c >= 10)

Then rc = 3+14r+10c

rc-14r-10c-3 = 0

rc - 14r - 10c +140 - 143 = 0

(r-10)(c-14) = 143

Now, the possibilities for (r,c) are (1,143), (11,13), (13,11) and (143,1). Checking all 4, the number of chairs in each is (11)(157), (21)(27), (23,25), (153,15).

Clearly (23)(25) > (21)(27) and (11)(157)<(153)(15). (if a>c>d>b>0 and a+b = c+d, ab < cd) Also, (153)(15) > (23)(25) obviously.

The maximum number of chairs is thus (153)(15) = 2295 (with 2292 chairs filled up)

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why ? defining X = r-10, Y = c-14, then that means X = 1,Y = 143 -> r = 11, c 157 -> rc = 1727 X = 11, Y = 13 -> r = 21, c = 27 -> rc = 567 X = 13, Y = 11 -> r = 23, c = 25 -> rc = 575 X = 143, Y = 1 -> r = 153, Y = 15 -> rc = 2295 (This is the max number of chairs)

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Why doesn't, say, r=21 and c=27 work?

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I hate it.. I hope they can fix it as soon as possible..

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There will be a lot to look forward to; new challenges and a huge database will be up tomorrow or Tuesday. Peter T. posted this earlier. I can't find a link at the moment, sorry!

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I got the same problem... Hope they fixed it soon.

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problems are back yepiii!!!!!!!!!!!!!!!!!!!!!

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They're up now.

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Why arent there any qns today? I thought they post qns here every Monday?!?!

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I hope they will bring new things.I will wait with patience

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:(

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But atleast they could send the weekly problems of one of the Olympiad sections. I'll get bored....

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Hopefully it's just because they're upgrading the techniques trainer.

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Hopefully

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i cant see my trignometry & calculus problems since 2 weeks...

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can you??

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