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AR15.COM
1/11/2008 10:33:26 AM EDT
I don't know how this question came into my head, but it's been driving me nuts since I can't figure it out.

Is there a number, and if so what is it, which is divisible by every whole number between 1 and 10  (inclusively) which will result in an integer?

_MaH

Edited to clarify (x2)

ETA - Just found out there is (using google - I cheated).  See if you can determine it on your own!

ETA 2 - Now that we've established there is, can you find the smallest number which will yield an accurate answer to the above question?
1/11/2008 10:35:01 AM EDT
[#1]
You want a number that can be divided by 1 through 10 and come out a whole number???



If so, then no.
1/11/2008 10:35:57 AM EDT
[#2]
I had a friend who divided by zero. Week later....he was dead!

True story.
1/11/2008 10:36:31 AM EDT
[#3]
Yes, there is.  Any number and every number, an infinity of numbers, can be divided by every number between one and ten.  Of course, there are also an infinity of numbers betweeen one and ten, as well.

I think you need to reformulate your question.  Were you perhaps referring to integer values only, or what?
1/11/2008 10:37:37 AM EDT
[#4]
there's only one thing you need to know about math:

pi
1/11/2008 10:37:39 AM EDT
[#5]

Quoted:
Yes, there is.  Any number and every number, an infinity of numbers, can be divided by every number between one and ten.  Of course, there are also an infinity of numbers betweeen one and ten, as well.

I think you need to reformulate your question.  Were you perhaps referring to integer values only, or what?


Yep.  Question clarified to reflect this.
1/11/2008 10:38:37 AM EDT
[#6]
In before Zaphod says 42.....

Oh, wait!
1/11/2008 10:39:11 AM EDT
[#7]
i had to write some stupid program to do this in computer science, to teach us how to use the modulus function..

ETA: here it is in C

#include <stdio.h>

int main () {
  int x=1;
  int stop=0;
  int divisor, remainder;

  while ( stop != 1 ) {
          divisor = 1;
          remainder = 0;
          while ( remainder == 0 && divisor < 10 ) {
                    divisor ++;
                    remainder = x % divisor;
           }        
           if ( remainder == 0 && divisor == 10 ) {
                printf("%i is divisible by 1 through 10", x);
                stop = 1;
           }
           else
                x ++;
  }

return 0;
}

ANSWER: 2520
1/11/2008 10:39:14 AM EDT
[#8]
3628800
1/11/2008 10:39:49 AM EDT
[#9]

Quoted:
I don't know how this question came into my head, but it's been driving me nuts since I can't figure it out.

Is there a number, and if so what is it, which is divisible by every number between 1 and 10  (inclusively) which will result in a whole integer?

_MaH

Edited to clarify


I am sure you meant to say "evenly" as in the result will be an integer.  

Yes, all you have to do to find that number is multiply all the numbers together that you want it to be evenly divisible by.

10!
1/11/2008 10:39:56 AM EDT
[#10]

Quoted:
I don't know how this question came into my head, but it's been driving me nuts since I can't figure it out.

Is there a number, and if so what is it, which is divisible by every number between 1 and 10  (inclusively) which will result in a whole integer?

_MaH

Edited to clarify


3.6288 million or 3,628,800

or 1x2x3x4x5x6x7x8x9x10

That is the easiest answer.

divide by 10, and you have 362,880 which is still divisible by 10, so that is an answer as well.
1/11/2008 10:40:55 AM EDT
[#11]
60,480
1/11/2008 10:41:02 AM EDT
[#12]
1/11/2008 10:41:45 AM EDT
[#13]

Quoted:

Quoted:
I don't know how this question came into my head, but it's been driving me nuts since I can't figure it out.

Is there a number, and if so what is it, which is divisible by every number between 1 and 10  (inclusively) which will result in a whole integer?

_MaH

Edited to clarify


I am sure you meant to say "evenly" as in the result will be an integer.  

Yes, all you have to do to find that number is multiply all the numbers together that you want it to be evenly divisible by.

10!



Neat....I didnt know that....
1/11/2008 10:44:04 AM EDT
[#14]

Quoted:
3816547290


Nope.  You used Google, didn't you?  'Cause I initially found that answer too


Quoted:
3628800



Quoted:
60,480


Yep!  It works!

Now can you find the SMALLEST number which will yield a correct answer?

_MaH
1/11/2008 10:44:35 AM EDT
[#15]
10!   (3,628,800)
1/11/2008 10:45:06 AM EDT
[#16]
5 x 7 x 8 x 9 = 2520

2520 / 1 = 2520
2520 / 2 = 1260
2520 / 3 =  840
2520 / 4 =  630
2520 / 5 =  504
2520 / 6 =  420
2520 / 7 =  360
2520 / 8 =  315
2520 / 9 =  280
2520 / 10 = 252
1/11/2008 10:46:29 AM EDT
[#17]

Quoted:
5 x 7 x 8 x 9 = 2520

2520 / 1 = 2520
2520 / 2 = 1260
2520 / 3 =  840
2520 / 4 =  630
2520 / 5 =  504
2520 / 6 =  420
2520 / 7 =  360
2520 / 8 =  315
2520 / 9 =  280
2520 / 10 = 252


Ladies and gentlemen, we have a winner for finding the smallest number!

Prize is bragging rights

_MaH
1/11/2008 10:47:47 AM EDT
[#18]
2520

ETA - damn it.  When I opened the thread - no answer.  Took me about 30 second.

Solution:

first go 1*2*3*4*5*6*7*8*9*10

Then factor out repeats to get

2(comes from requirement to be divis by 8)*3*3(this three comes from divisibility by 9)*4*5
1/11/2008 10:47:48 AM EDT
[#19]
There's probably lots of such numbers.

The lowest one I come up with is 7,560.

See, you don't need to 10!, because 6 is just 2x3, 8 is 2x4, and 10 is 2x5.

So the low answer is 2x3x4x5x7x9.

All you need to do is multiply the numbers in the series that aren't factors of each other for your answer.

So, what do I win??


ETA:  Scooped (and beaten) whilst typing my reply.
1/11/2008 10:56:09 AM EDT
[#20]

Quoted:
There's probably lots of such numbers.

The lowest one I come up with is 7,560.

See, you don't need to 10!, because 6 is just 2x3, 8 is 2x4, and 10 is 2x5.

So the low answer is 2x3x4x5x7x9.

All you need to do is multiply the numbers in the series that aren't factors of each other for your answer.

So, what do I win??


ETA:  Scooped (and beaten) whilst typing my reply.


You don't need 2 because 4 is already 2x2.  You don't need 3 either, since 9 = 3x3.

You do need 8 though, and 6 as well, I think.
1/11/2008 11:04:43 AM EDT
[#21]

Quoted:

Quoted:
There's probably lots of such numbers.

The lowest one I come up with is 7,560.

See, you don't need to 10!, because 6 is just 2x3, 8 is 2x4, and 10 is 2x5.

So the low answer is 2x3x4x5x7x9.

All you need to do is multiply the numbers in the series that aren't factors of each other for your answer.

So, what do I win??


ETA:  Scooped (and beaten) whilst typing my reply.


You don't need 2 because 4 is already 2x2.  You don't need 3 either, since 9 = 3x3.

You do need 8 though, and 6 as well, I think.


Negative.  8 = 2*2*2, 6 = 2*3

Really the only numbers you need are the primes (2, 3, 5, 7).

If you breakdown 1 - 10 by their prime factors, and multiply the prime factors to the highest power, it'll look like this:

1 = 1*1
2 = 1*2
3 = 1*3
4 = 2*2
5 = 1*5
6 = 2*3
7 = 1*7
8 = 2*2*2
9 = 3*3
10 = 2*5

(2^3) * (3^2) * (5^1) * (7^1) =
8 * 9 * 5 * 7 =
72 * 35 = 2520

_MaH
1/11/2008 11:13:51 AM EDT
[#22]

Quoted:
there's only one thing you need to know about math:

pi


I like pi. Pi is wonderful... If you're really lucky you'll find a woman that likes pi too...
1/11/2008 11:23:00 AM EDT
[#23]

Quoted:

Quoted:
I don't know how this question came into my head, but it's been driving me nuts since I can't figure it out.

Is there a number, and if so what is it, which is divisible by every number between 1 and 10  (inclusively) which will result in a whole integer?

_MaH

Edited to clarify


3.6288 million or 3,628,800

or 1x2x3x4x5x6x7x8x9x10

That is the easiest answer.

divide by 10, and you have 362,880 which is still divisible by 10, so that is an answer as well.


362,880 will work too.

15,120
1/11/2008 12:58:02 PM EDT
[#24]

Quoted:
Negative.  8 = 2*2*2, 6 = 2*3

Really the only numbers you need are the primes (2, 3, 5, 7).

If you breakdown 1 - 10 by their prime factors, and multiply the prime factors to the highest power, it'll look like this:

1 = 1*1
2 = 1*2
3 = 1*3
4 = 2*2
5 = 1*5
6 = 2*3
7 = 1*7
8 = 2*2*2
9 = 3*3
10 = 2*5

(2^3) * (3^2) * (5^1) * (7^1) =
8 * 9 * 5 * 7 =
72 * 35 = 2520

_MaH


Nice work, mhoffman!  
1/11/2008 2:58:39 PM EDT
[#25]

Quoted:

Quoted:
Negative.  8 = 2*2*2, 6 = 2*3

Really the only numbers you need are the primes (2, 3, 5, 7).

If you breakdown 1 - 10 by their prime factors, and multiply the prime factors to the highest power, it'll look like this:

1 = 1*1
2 = 1*2
3 = 1*3
4 = 2*2
5 = 1*5
6 = 2*3
7 = 1*7
8 = 2*2*2
9 = 3*3
10 = 2*5

(you jumped over a whole bunch of steps that go right here to get to your answer)


(2^3) * (3^2) * (5^1) * (7^1) =
8 * 9 * 5 * 7 =
72 * 35 = 2520

_MaH


Nice work, mhoffman!  


Show me a mathematical proof that this is correct and complete.
1/11/2008 3:02:45 PM EDT
[#26]
The first one I came up with was 630 x 4, or 2,520.

(630 = 70x9 = 7x90 )
1/11/2008 4:35:33 PM EDT
[#27]
This is just finding LCM. Prime numbers is the key. Any whole number except 0 and 1 can be represented as multiples of prime numbers.
so between 1 and 10, 2, 3, 5, 7 are the ones, so find combination of each prime numbers. The key is to find the most number of times each prime number shows up.

8 is 2*2*2 it has to be the one that shows up for 2.

For 3, it is 3*3 and 5 and 7 comes out once, so the result is 2*2*2*3*3*5*7 = 2520.

edit: doh! mhoffman did it FYI any other number that satisfies this requirement will be multiples of 2520