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Numerous Numeral Enumeration (Posted on 2011-03-20) Difficulty: 3 of 5
(A) Determine the total number of ways in which 102010(base ten) is expressible as the product of:

(I) Four distinct positive integers arranged in increasing order of magnitude.

(II) Five distinct positive integers arranged in increasing order of magnitude.

(III) Six distinct positive integers arranged in increasing order of magnitude.

(B) What are the respective answers to each of (I), (II) and (III) in part-(A), if 102010(base ten) is replaced by 102010(base 12)?

No Solution Yet Submitted by K Sengupta    
Rating: 4.5000 (2 votes)

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Some Thoughts re: Initial thoughts. | Comment 4 of 10 |
(In reply to Initial thoughts. by Jer)

Still attempting only part I:

We can consider the exponents of 5, and the exponents of 2 separately until we later need to combine them.

We can use the same terminology as in the solution to  "So many quintuplets".

For a given set of numbers, 0 through 2010, that add up to 2010 the representation is 4 x's interspersed among 2010 + signs if starting from zero. That's C(2014,4) = 683,489,813,501, which would be the answer if the base were a prime number, such as 2 or 5, rather than the 10 that it is.

The same would apply individually to the power set for 2 as for 5. Together, you'd get the product of these two: 467,158,325,159,631,761,877,001. But that assumes that if the powers of 5 were laid out in ascending sequence and so were the powers of 2, you'd match them up in the same order. That's not the case. You could match them up in 4! ways, so the number needs to be multiplied by 4! = 24, resulting in 11,211,799,803,831,162,285,048,024.

That's an overestimate, but I don't think by much. That's because the ones needing elimination are those where there's at least one duplicate among the powers of 5 and also at least one duplicate among the powers of 2, and that at least one of those duplications on each side (2 or 5) is the same as one of the duplications on the other side.

To see how many non-duplicated sets of numbers add up to 2010 we tie each x in the representation to a +, say on its left, so that +x's mix with +'s. Since we can still start with zero, the initial number in our heads has to be -1, which is immediately changed to zero, via either a + or a +x. There are 4 +x's and 2007 lone +'s (remembering we have to get to 2010 from -1), so the number is C(2011,4) = 679,423,448,830 sets that have no duplicate powers of 5 (or of 2 when doing the 2's).

So in either set (2's or 5's) only 683,489,813,501 - 679,423,448,830 = 4,066,364,671 have any duplicates. In combination, only 4,066,364,671 * 4,066,364,671 = 16,535,321,637,556,938,241 have duplications for both the powers of 5 and the powers of 2. That sounds like a lot, but it's out of 11,211,799,803,831,162,285,048,024, or 1/678051.5099485953512807292....

Not only that, but during the shuffling of the powers of 2 to match with the powers of 5, numbers that matched on the one side would have to be numbers that matched on the other, so as to get, for example 5^172*2^85 twice, when 172 appears twice in the 5's list and 85 appears twice in the 2's list.

To do the calculation properly, you'd need to take into consideration the probability of one or two pair, or 3-of-a-kind in either or both of the powers-of-5 list and the powers-of-2 list, and determine the probabilities there, and produce and combine the actual numbers(from the probabilities).

It only gets more complicated for the other cases, where you'd have "full houses", two threes-of-kind, three pairs, etc.

  Posted by Charlie on 2011-03-21 10:37:04
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