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 April 5th, 2007, 07:18 AM #1 Member   Joined: Nov 2006 Posts: 54 Thanks: 0 A Decimal Digit Puzzle Determine the total number of non-negative integers with not more than 1993 decimal digits having non-decreasing digits. For example, 33455 is a valid instance of such non-negative integers, while 76 is NOT a valid non-negative integer having non-decreasing digits.
 April 5th, 2007, 12:40 PM #2 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Answer > 1993
 April 5th, 2007, 06:24 PM #3 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 Hey, balls and urns argument! We have 1993 identical objects and nine dividers. The objects before the first divider are 0's, those between the first and second are 1's, etc. We now have 1993+9=2002 total objects, which fall into two categories, the objects in one category being indistinguishable from each other. Therefore, there are 2002!\(1993!9!)≈1.39827649874513*10^24 such numbers. The last three digits of the number should be zeros, so that means we are only three digits shy of having an exact answer (my calculator only uses 15 significant digits). But CRGreathouse should have no problem giving the exact answer!
April 6th, 2007, 03:28 PM   #4
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Quote:
 Originally Posted by roadnottaken The last three digits of the number should be zeros, so that means we are only three digits shy of having an exact answer (my calculator only uses 15 significant digits). But CRGreathouse should have no problem giving the exact answer!
Well,the last two anyhow; you run out of spare 2s faster than 5s in this case. It's 1398276498745133921413500.

 April 6th, 2007, 06:43 PM #5 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 Hmm, I thought I checked to make sure it had enough 2's, but apparently not. Fortunately, no one caught my mis-subtraction (24-15=9=/=6 [I must have been up too late]).

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