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Gross Division (Posted on 2006-02-20) Difficulty: 3 of 5
Find the smallest positive integer n such that n has exactly 144 distinct positive divisors and there are 10 consecutive integers among them (Note: 1 and n are both divisors of n)

See The Solution Submitted by goFish    
Rating: 3.8000 (5 votes)

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Solution Slightly improved | Comment 4 of 14 |

I looked at Steve's comment, so that's where I got my ideas.

I start with the smallest 10 consecutive integers possible:
1,2,3,4,5,6,7,8,9,10

The factors required for these divisors:
2^3*3^2*5*7

Now I have to add factors to get 144 divisors.  Note that the number of divisors is equal to the product of each exponent plus 1.  So there are 4*3*2*2 divisors in the number 2^3*3^2*5*7.  Possible factors to add are the following:
2^8 = 256
2^2*5^2 = 100
3*5*7 = 105
2^2*3*5 = 60

3*5*7 appears to lead to Steve's solution.  But 2^2*3*5 will improve it. It may possibly be the best solution, but I'm not making any promises.

So, the number is 2^5*3^3*5^2*7 = 151200


  Posted by Tristan on 2006-02-20 12:09:29
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