General Question

mattbrowne's avatar

What is your favorite math problem?

Asked by mattbrowne (31719points) June 21st, 2009

Here’s one: it’s Christmas Eve, Jesus’s birthday. There are 365 people at church. The minister asks: Anyone else celebrating your birthday today? Raise your hands. No hands. Hmm. Is this very unlikely? (assumption: there are no shy people in this church)

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22 Answers

Jayne's avatar

@mattbrowne; 00110110 00110011 00101110 00110010 00110110 00100101 00100000 01110100 01101000 01100001 01110100 00100000 01110011 01101111 01101101 01100101 01101111 01101110 01100101 00100000 01100100 01101111 01100101 01110011 00101100

SuperMouse's avatar

I am a fan of the quadratic equation.

buster's avatar

Why was 6 scared of 7? 7,8,9.

SirBailey's avatar

I like Pi. Especially apple.

whitenoise's avatar

If a year is 365 days and one rotation around the sun, then how often does the earth rotate per year?

(You need some additional research, I guess.)

PandoraBoxx's avatar

There are 1,000 lockers in the long hall of Westfalls High. In preparation for the beginning of school, the janitor cleans the lockers and paints fresh numbers on the locker doors. The lockers are numbered from 1 to 1,000. When the 1,000 Westfalls High students return from summer vacation, they decide to celebrate the beginning of the school year by working off some energy.

* Student 1 opens every locker.
* Student 2 either opens or closes every other locker.
* Student 3 opens or closes every third locker. And so on, ...

Which locker doors are open when every student finishes?

whitenoise's avatar

you mean which 31 lockers are open, right?

mammal's avatar

any division by zero
is problematic

Clair's avatar

Ever since the first grade, I’ve always been obsessed with the Fibonacci sequence. Not sure why.
Not really a problem, I know.

jfos's avatar

@mattbrowne I think you have the problem wrong… There is no guarantee that anybody has the birthday on Christmas eve, as they could all have their birthdays on March 6.

The trick is that, mathmatically, AT LEAST two people share the same birthday.

jfos's avatar

i.e. 365 people plus the minister—> 366 birthdays.

whitenoise's avatar

@jfos That is in a world without 29 February, you mean. ;-)

mattbrowne's avatar

@jfos – Of course there’s no guarantee. I asked: Is this very unlikely? Can you tell?

mattbrowne's avatar

@Jayne – My solution looks like this: 00110110 00110010 00101110 00110011 00110100 00100101 00100000 01110100 01101000 01100001 01110100 00100000 01110011 01101111 01101101 01100101 01101111 01101110 01100101 00100000 01100100 01101111 01100101 01110011 00101100 or simply 31 2d 28 33 36 34 2f 33 36 35 29 5e 33 35 36 3d

Jayne's avatar

@mattbrowne; why ^356 instead of ^365? Neither Jesus nor the minister should be included (unless the minister is really weird and raises his hand for his own questions)

mattbrowne's avatar

@Jayne – Yes, there was a typo. Sorry about that. Of course it’s 1—(364/365)^365 (let’s not consider the issue of leap years).

It’s actually a real story which happened when my kids were about 15. After church they asked me, “Isn’t it weird that it’s no one’s birthday today? The church was packed with almost 400 people.” I replied that my guess is there’s one chance in three something like that happens.” “No way, Dad.” “Let’s get our pocket calculator. Or let’s just use Google.” “Google doesn’t know things like that.” “I’ll show you.” and we typed (364/365)^365 into the search field. They were not aware that Google was a calculator. Normally they know more about the web than I do. The result surprised them.

It’s my opinion that real life problems like this is what gets kids interested in math. Otherwise it’s just a set of boring formulas on black boards.

We need to share our favorite math problems with kids and get them interested in science and technology.

Here’s another one: You’ve got a toy cannon shooting balls across the room. When the cannon is on the floor 45 degrees is the optimal angle to shoot as far as possible (we ignore air, wind etc.

Now you got a table which is h cm high. Tell me the optimal angle depending on h. It’s obvious that the angle gets smaller the larger h gets. But what is it exactly? I remember discussion this in physics class in 11th grade.

Noel_S_Leitmotiv's avatar

Noel has six beers. He wants to share them with a good friend. How many will his friend get if Noel gives his friend half of them.

ragingloli's avatar

he gets nothing, because if you cut the cans in half, the beer will spill and he would end up with empty husks.

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