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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)

  Submitted by goFish    
Rating: 3.8000 (5 votes)
Solution: (Hide)
The smallest sequence is 1 ... 10 which has lowest common multiple of 2,520 and we note that

2520 = (2^3)(3^2)(5^1)(7^1).

Taking the exponents (3, 2, 1, 1), adding 1 to each gives (4, 3, 2, 2) which when multiplied gives 48 as the number of divisors of 2520. We now look to add to the exponents so that the same product gives 144.

Adding a new prime the product of (6,3,2,2,2) gives 144. Reducing each by 1 gives (5,2,1,1,1) which when attached to the primes gives (2^5)(3^2)(5^1)(7^1)(11^1) = 110,880 as the smallest n.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(2): I think I got itChelsea2006-02-21 13:19:06
re: I think I got it (spoiler?) ...yes, you didAdy TZIDON2006-02-21 08:43:15
ugotitAdy TZIDON2006-02-21 08:41:14
re: I think I got it (spoiler?)Mindrod2006-02-20 22:38:43
Solutioncomputer solutionCharlie2006-02-20 17:35:30
I think I got it (spoiler?)Chelsea2006-02-20 16:29:58
re: not any lower..........spoiler? NogoFish2006-02-20 16:04:37
Solutionnot any lower..........spoilerAdy TZIDON2006-02-20 15:50:42
LowergoFish2006-02-20 14:51:54
re: improved furtherMindrod2006-02-20 14:24:52
SolutionSlightly improvedTristan2006-02-20 12:09:29
re: startCharlie2006-02-20 10:34:58
Some ThoughtsstartCharlie2006-02-20 10:32:24
Some ThoughtsSpoiler? Possible Solution?Steve Herman2006-02-20 10:28:20
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