A math problem for you geeks out there.

[Fireman]

My 5th grader brought this home on an assignment.
Post your answer and how you figured it out.

Q: All whole numbers from 1 to 2006 are written on on a black board. Johnny underlines all numbers divisible by 2. Sara underlines all numbers divisible by 3. Rudy underlines all numbers divisible by 4.
How many numbers are underlined exactly twice?

 

[reverend boltron]

668.

Edit:
I divided 2006 by four and I got 501.5, and I knew that all of those numbers had to be divisible by two.

So those were underlined exactly twice.

Then I took 2006 and divided it by 3, and I got 668. I divided that by two, because I knew that only half of those would be even, which is a criteria to be underlined by 2 or 4.

Then I realized that every number that was divisible by 4 and 3 was also divisible by 2, which was 3 underlines. So I subtracted every number that was divisible by 12 in the range of 2006. So 2006/12= 167.

So, 501+334-167 = 668

Edit 2: I didn't "subtract" the numbers that were divisible by 12, but I removed them from the list.

 

[Fireman]

OK but how did you get it?

 

[chrisms]

Divide 2006 by two. Divide that number by two, thats how many are divisible by 2 and 4. Then every other number divisible by 3 is even, probably some pattern with the 3 and 4 numbers. Too lazy to think of it.

 

[Fireman]

'I divided 2006 by four and I got 501.5, and I knew that all of those numbers had to be divisible by two.

So those were underlined exactly twice.
'

This is wrong because some of those numbers will also be divisible by 3 and will be underlined 3 times.

 

[reverend boltron]

Originally posted by: Fireman


'I divided 2006 by four and I got 501.5, and I knew that all of those numbers had to be divisible by two.

So those were underlined exactly twice.
'

This is wrong because some of those numbers will also be divisible by 3 and will be underlined 3 times.

Then take out all of the ones that are divisible by six and you have your answer.

If you're taking out the ones that are divisible by six, then don't take out the ones that are divisible by twelve too. Then you're coming up short.

So I guess it is 501.

 

[MartyMcFly3]

Its Saturday night. I dont challenge my brain on saturday night. If anything I slowly kill it, attacking it one brain cell at a time.

 

[Fireman]

"Then take out all of the ones that are divisible by six and you have your answer.

If you're taking out the ones that are divisible by six, then don't take out the ones that are divisible by twelve too. Then you're coming up short.

So I guess it is 501. "

Winner winner chicken dinner! 501
Took me a lot longer though. Good job!

 

[KLin]

Fix(2006/4) = answer

 

[nd]

I got 501

 

[KLin]

I'm so bored I wrote an excel macro to evaluate if there were exactly 2 whole numbers for each divisible number. . Is Pan's Labyrinth any good?

 

[reverend boltron]

Thanks! I like the problem of Hotel Infinity.

You have a hotel where there are an infinite amount of rooms, but managing an infinite amount of rooms is an awfully hard job to do. So they go through managers about once a month. The first manager of Hotel Infinity had every other room as a smoking room. But when he left, the next manager switched the status of every third room. And after he left, the manager after him switched the status of every fourth room. The question is, after an infinite amount of managers, what rooms are smoking?

 

[Mr Smiley]

Im so bored I wrote a php script to answer it for me.
<?php
$numbers = 0;
$i = 1;
while($i<=2006){
if($i%2==0 && $i%3==0 && $i%4!=0){
$numbers++;
}
elseif($i%2==0 && $i%3!=0 && $i%4==0){
$numbers++;
}
elseif($i%2!=0 && $i%3==0 && $i%4==0){
$numbers++;
}
$i++;
}
echo $numbers;
?>
Heh 501!

 

[ungsunghero]

Originally posted by: reverend boltron
668.

Edit:
I divided 2006 by four and I got 501.5, and I knew that all of those numbers had to be divisible by two.

So those were underlined exactly twice.

Then I took 2006 and divided it by 3, and I got 668. I divided that by two, because I knew that only half of those would be even, which is a criteria to be underlined by 2 or 4.

Then I realized that every number that was divisible by 4 and 3 was also divisible by 2, which was 3 underlines. So I subtracted every number that was divisible by 12 in the range of 2006. So 2006/12= 167.

So, 501+334-167 = 668

Edit 2: I didn't "subtract" the numbers that were divisible by 12, but I removed them from the list.

I did pretty much the same as this.

I looked for all numbers divisible by 4 or 6, and not divisible by 12 (which would be those numbers with three underlines).

501 numbers are divisible by 4, 1/3 of them (167) by 12.
334 numbers are divisible by 6, 1/2 of them (167) by 12.

501 + 334 - 167 - 167 = 501

At first, I was looking for numbers divisble by 6 and 8, but not 12, but that would have left out those numbers divisble by 4 and not 3 (4, 16, 28, etc.)

 

[KLin]

bah code looks stupid in OT.

 

[gsethi]

edit: nm...

 

[KLin]

Since my brain is too bored to work correctly, why is 2006/4 and dropping the remainder the answer?

 

[JujuFish]

Originally posted by: KLin
Since my brain is too bored to work correctly, why is 2006/4 and dropping the remainder the answer?

Because the third of those which are divisible by 4 is exactly equal to the amount that are divisible by 2 and 3 and not 4, so they cancel each other out.

 

[Scarpozzi]

This is one of many reasons I don't want to have kids. If those snotty little bastards bring me meaningless problems to do, I think I'm going to abort them.....no matter how old they happen to be at the time. :disgust:

 

[Adn4n]

Every number is divisible by any number other than 0, thus, all are underlined.

 

[eLiu]

Here's a similar problem...

Say all the numbers from 1 to 2006 are written on the board. Then the students underline each number that is divisible by 1, divisible by 2, divisible by 3, and so on, all the way up to all numbers divisible by 2006. How many numbers are underlined an *odd* number of times?

 

[b0mbrman]

Originally posted by: Fireman
My 5th grader brought this home on an assignment.
Post your answer and how you figured it out.

Q: All whole numbers from 1 to 2006 are written on on a black board. Johnny underlines all numbers divisible by 2. Sara underlines all numbers divisible by 3. Rudy underlines all numbers divisible by 4.
How many numbers are underlined exactly twice?

Cycles at 12.

From 1 to 12, the number 4, 6, 8 are underlined twice.

167 cycles plus 2.

167 x 3 = 501

 

[FilmCamera]

The correct answer is who cares?

 

[XZeroII]

Originally posted by: reverend boltron
Thanks! I like the problem of Hotel Infinity.

You have a hotel where there are an infinite amount of rooms, but managing an infinite amount of rooms is an awfully hard job to do. So they go through managers about once a month. The first manager of Hotel Infinity had every other room as a smoking room. But when he left, the next manager switched the status of every third room. And after he left, the manager after him switched the status of every fourth room. The question is, after an infinite amount of managers, what rooms are smoking?

An infinite number.

 

[XZeroII]

Originally posted by: Fireman
My 5th grader brought this home on an assignment.
Post your answer and how you figured it out.

Q: All whole numbers from 1 to 2006 are written on on a black board. Johnny underlines all numbers divisible by 2. Sara underlines all numbers divisible by 3. Rudy underlines all numbers divisible by 4.
How many numbers are underlined exactly twice?

42.

In 5th grade, I really didn't care enough to put more than 2 seconds worth of thought into a single questions.