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Topic: 21 FACTORIAL (Read 4482 times) |
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jonderry
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21 FACTORIAL
« on: Aug 18th, 2004, 2:04pm » |
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I can't think of an elegant way to do this.
We know that x + y = 2 (mod 9) which gives
the possible combinations (0,2), (1,1), (2,0), (2,9), (3,8 ), (4,7), (5,6), (6,5), (7,4), (8,3), and (9,2).
That's all I can get without serious number crunching.
Does anyone know how to do this, and if so, can you give me a hint?
Strange that this is in "easy" and 100 factorial is in "hard" unless I am missing something obvious.
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| « Last Edit: Aug 18th, 2004, 2:06pm by jonderry » |
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Grimbal
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Re: 21 FACTORIAL
« Reply #1 on: Aug 18th, 2004, 2:19pm » |
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It is also a multiple of eleven.
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jonderry
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Re: 21 FACTORIAL
« Reply #2 on: Aug 18th, 2004, 5:37pm » |
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I don't know of any interesting properties of multiples of 11. What property of multiples of 11 is useful for quickly solving this problem?
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Icarus
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Re: 21 FACTORIAL
« Reply #3 on: Aug 18th, 2004, 8:12pm » |
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If n is even, then 11 divides 10n - 1. If n is odd, 11 divides 10n+1. From these two facts, you can prove that for any number, the difference between the sum of its digits in even position and the sum of its digits in odd position is congruent to the number itself modulo 11:
If x = [sum]i xi10i, then
x [equiv] [sum]i x2i - [sum]i x2i+1 mod 11.
Thus, for example, 12345678 [equiv] ( 2+4+6+8 ) - ( 1+3+5+7 ) [equiv] 4 mod 11.
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| « Last Edit: Aug 18th, 2004, 8:23pm by Icarus » |
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jonderry
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Re: 21 FACTORIAL
« Reply #4 on: Aug 19th, 2004, 10:56am » |
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I see. And that property can also be proved by noting that adding a 1 to an even (odd) digit always increases the tally mod 11 of the even (odd) columns by 1 relative to the odds (evens) (even when there is a possibly cascading carry). Thus, adding 11 always keeps the tallies equal modulo 11.
So the answer is:
x=4,y=7
right?
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| « Last Edit: Aug 19th, 2004, 11:06am by jonderry » |
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soundar
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this is an elementary question.
we know 21! has 9 and 11 as factors; hence the sum of the digits must be divisible by 9 and the difference of sums of odd-place digits and even-place digits must be divisible by 11.
hence we have x+y=2 mod 9; or, x+y=2 or 11
also, (41+y)-(11+x)=0 mod 11; or y-x=-8 mod 11; or,
y-x=-8 or 3
so we have four sets of equations, 1) x+y=2, y-x=-8
2) x+y=2, y-x=3 3) x+y=11, y-x=-8 4) x+y=11, y-x=3
the first set results in 2y=-6, which is not possible; the second set results in 2y=5 and the third in 2y=3, both not possible
hence the last set must give us the solution, 2y=14, or y=7 and hence x=4
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Icarus
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Re: 21 FACTORIAL
« Reply #6 on: Sep 19th, 2004, 8:40pm » |
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It is elementary only if you are familiar with the rules for determining divisability by 9 & 11. Jonderry had never come across the "divisable by 11" rule.
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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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murdok03
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Re: 21 FACTORIAL
« Reply #7 on: Oct 1st, 2012, 7:25am » |
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I spent so much time on this that I feel I have to post my solution. Even though it's not the most elegant.
1. the number has to be dividable by 9
2. the number has to be dividable by 7
3. x & y single digits( 0...9 )
1. the sum of all digits has to be dividable by 9:
=> x+y=2 or x+y=11
2. the number has to be dividable by 7:
see: h t t p : // math.about.com/library/bldivide.htm
( I had to google this one )
it's complicated but:
3x+4y={5,12,19,26,33,40,47,54}
if we solve for x+y=2 none have solutions
if we solve for x+y=11
y+33={5,12,19,26,33,40,47,54}
where y={0..9} => y+33=40 => y=7
and x=11-7=4
And than I found this forum, hmm too bad I didn't think of divide by 11 it would of been alot simpler
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JudyNeary
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Re: 21 FACTORIAL
« Reply #8 on: Oct 8th, 2012, 10:37pm » |
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Hi ! you simply this coding using c. i hope you get the result perfectly.
#include <stdio.h>
int main()
{
int c, n=21, fact = 1;
printf("Enter a number to calculate it's factorial\n");
scanf("%d", &n);
for (c = 1; c <= n; c++)
fact = fact * c;
printf("Factorial of %d = %d\n", n, fact);
return 0;
}
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towr
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Re: 21 FACTORIAL
« Reply #9 on: Oct 9th, 2012, 8:52am » |
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The problem is NOT to calculate 21 factorial, but to solve the puzzle without doing exactly that:
Quote:21!=510909x21y1709440000
Without calculating 21!, what are the digits marked x and y? |
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Wikipedia, Google, Mathworld, Integer sequence DB
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