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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Initial thoughts. | Comment 5 of 10 |
(In reply to re: Initial thoughts. by Charlie)

I disagree with three things in this:

"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."

If anything it would be C(2010,3).  There are 2010 + signs, you are choosing which three to change to *, which would create your four exponents.
It doesnt account for numbers in increasing order of magnitude.
It also doesnt account for keeping the numbers distinct.

Using a smaller example: 2^8  can only be broken into four distinct positive integers arranged in increasing order of magnitude in two ways:
1,2,4,32 and
1,2,8,16.



  Posted by Jer on 2011-03-21 14:07:59

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