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Topic: 21! (Read 2422 times) |
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BNC
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21!=510909x21y1709440000
Without calculating 21!, what are the digits marked x and y?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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NickH
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Re: 21!
« Reply #1 on: Apr 2nd, 2003, 12:42pm » |
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Solution below...
x+y = 2 (mod 9), since 21! = 0 (mod 9.)
So x+y = 2 or 11.
x-y = 8 (mod 11), since 21! = 0 (mod 11.)
So x-y = 8 or -3.
x+y = 2 is easily ruled out by the mod 11 constraints.
x+y = 11, x-y = 8 has fractional solutions.
x+y = 11, x-y = -3 has solutions x = 4, y = 7.
(And yes, I did check the solution with a calculator, but only after I'd derived it!)
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| « Last Edit: Apr 2nd, 2003, 12:43pm by NickH » |
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Nick's Mathematical Puzzles
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Icarus
wu::riddles Moderator
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Boldly going where even angels fear to tread.
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Re: 21!
« Reply #2 on: Apr 2nd, 2003, 3:40pm » |
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A note of explanation about Nick's solution, for those that are not already familiar with it:
The notation "x = y mod n" (read "x is congruent to y modulo n") means that n evenly divides (x-y). That is, x and y differ by a multiple of n. (Usually there is a third line in the "equal" sign, but this isn't available in the average keyboard.)
Nick gets his first formula from the following fact:
every decimal number is congruent modulo 9 to the sum of its digits (this is a slightly stronger version of testing divisibility by 9 by summing the digits). For example,
4753 = 4+7+5+3 = 19 = 1+9 = 10 = 1+0 = 1 mod 9.
His second formula comes from the less well-known result:
every decimal number is congruent modulo 11 to (the sum of the digits in odd positions) minus (the sum of the digits in even positions) For example
4753 = (3 + 7) - (5 + 4) = 1 mod 11.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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