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Divides them all.... (Posted on 2010-11-11) Difficulty: 2 of 5
What is the smallest positive number that is evenly divisible by each of the integers from 1 to 24 -inclusive?

No Solution Yet Submitted by Ady TZIDON    
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Solution? | Comment 1 of 2
Clearly all of the primes between 1 and 24 will need to be factors of this number, so let's start with:

2*3*5*7*11*13*17*19*23 = 223092870

Now we just need to check the composite numbers between 1 and 24 to see if they can be formed by multiplying any combination of these primes. If not, we need to add the smallest factor that will make this possible:

4 - no (add a second 2 as a factor)
6 - yes (2*3)
8 - no (add a third 2 as a factor)
9 - no (add a second 3 as a factor)
10 - yes (2*5)
12 - yes (2*2*3)
14 - yes (2*7)
15 - yes (3*5)
16 - no (add a fourth 2 as a factor)
18 - yes (2*3*3)
20 - yes (2*2*5)
21 - yes (3*7)
22 - yes (2*11)
24 - yes (2*2*2*3)

So our final number is 2*2*2*2*3*3*5*7*11*13*17*19*23, or 223092870*24 = 5354228880.

At this point it also becomes fairly easy to keep going, and find the smallest number evenly divisible by all integers between 1 and 25, 1 and 26, etc.:

(1-25, 1-26): 5354228880*5 = 26771144400
(1-27, 1-28) = 2677114440*3 = 8031343320
(1-29, 1-30) = 8031343320*29 = 2329089562800
(1-31, 1-32, 1-33, 1-34, 1-35, 1-36) = 2329089562800*31 = 72201776446800

In general, moving forward, our number requires a new factor only when we wish to make it divisible by a new prime or a prime raised to a power greater than 1.

Edited on November 11, 2010, 11:55 am
  Posted by Jyqm on 2010-11-11 11:43:45

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