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 Gross Division (Posted on 2006-02-20)
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.

 Subject Author Date re(2): I think I got it Chelsea 2006-02-21 13:19:06 re: I think I got it (spoiler?) ...yes, you did Ady TZIDON 2006-02-21 08:43:15 ugotit Ady TZIDON 2006-02-21 08:41:14 re: I think I got it (spoiler?) Mindrod 2006-02-20 22:38:43 computer solution Charlie 2006-02-20 17:35:30 I think I got it (spoiler?) Chelsea 2006-02-20 16:29:58 re: not any lower..........spoiler? No goFish 2006-02-20 16:04:37 not any lower..........spoiler Ady TZIDON 2006-02-20 15:50:42 Lower goFish 2006-02-20 14:51:54 re: improved further Mindrod 2006-02-20 14:24:52 Slightly improved Tristan 2006-02-20 12:09:29 re: start Charlie 2006-02-20 10:34:58 start Charlie 2006-02-20 10:32:24 Spoiler? Possible Solution? Steve Herman 2006-02-20 10:28:20

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