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Topic: MS question 3 (Read 10652 times) |
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kiki lee
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Generate a 9 digit number using the number 1-9, without using any number
more than once. this number must also be divisible such that if you take
the left most digit the number is divisible by 1, if you take the two
left most digits the number is divisible by 2, if you take the three left
most digits the number is divisible by 3, etc., etc. What is the number?
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Ozzie
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Can you please clarify what you mean by take the left n digits ? Does that mean, for example, that if you are taking 3 left digits, that you are trying to make the 3 digits divisible by 3 or the remaining 6 digit number divisible by 3 ?
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towr
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Re: MS question 3
« Reply #2 on: Jun 26th, 2003, 11:49am » |
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I think she means you have number a,b,c,d,e,f,g,h,i that together are the numbers 1-9
and
i % 1 = 0 (% = modulus)
hi % 2 = 0
ghi % 3 = 0
fghi % 4 = 0
efghi % 5 = 0
defghi % 6 = 0
cdefghi % 7 = 0
bcdefghi % 8 = 0
abcdeghi % 9 = 0
I think it's also allready on the site..
In any case it's easy to brute-force it with a simple program (trying every combination)
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| « Last Edit: Jun 26th, 2003, 11:51am by towr » |
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wowbagger
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Re: MS question 3
« Reply #3 on: Jun 26th, 2003, 11:54am » |
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I almost agree with you, towr.
You do know left from right, don't you? 
Maybe someone like BNC will come up with a non-brute force answer.
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BNC
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Re: MS question 3
« Reply #4 on: Jun 26th, 2003, 11:54am » |
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on Jun 26th, 2003, 11:49am, towr wrote:
I think it's also allready on the site..
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here
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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towr
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Re: MS question 3
« Reply #5 on: Jun 27th, 2003, 1:11am » |
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on Jun 26th, 2003, 11:54am, wowbagger wrote:I almost agree with you, towr.
You do know left from right, don't you? |
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Not usually.. why? 
heh, I did get it right in the other thread 
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| « Last Edit: Jun 27th, 2003, 1:13am by towr » |
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Claire Main
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Actually the question is slightly different...this question asks for a 9 digit number using the digits 1-9, the other asks for a 10-digit number using 0-9 
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Icarus
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Re: MS question 3
« Reply #7 on: Sep 17th, 2004, 7:12pm » |
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True - but if you can find the answer for one, the other is fairly obvious, isn't it?
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Grimbal
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Re: MS question 3
« Reply #8 on: Sep 18th, 2004, 11:19am » |
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We can rephrase the conditions
(note a|b means a divides b i.e. b = a*n)
1. always true
2. 2|b
3. 3|a+b+c
4. 4|cd
5. e = 5, and no other digit if 5
6. 2|f and 3|d+e+f
7. 7|abcdefg
8. 8|gh (8|fgh, but f is even)
9. always true if all digits are used
- b,d,f,h are even, so a,c,e,g,i are odd.
- d+e+f must be an odd multiple of 3 with e=5. Only 258, 456, 654 and 852 are possible for def.
- 4|cd, and c odd implies d is 2 or 6. def = 258 or 654.
- a+b+c and g+h+i are even and multiples of 3.
- 8|gh and g odd implies g is 2 or 6. gh = 16, 32, 72, 96. (56 uses 5).
- 6|g+h+i, implies ghi is one of 321, 327, 723, 729, 963.
- the only possible def are resp. 654, 654, 654, 654, 258.
- the only possible abc or cba are 789, 189, 189, 183, 147.
This leaves only the following numbers
789654321
987654321
189654327
981654327
189654723
981654723
183654729
381654729
741258963
147258963
they all satisfy all divisibilities except by 7.
Only 381654729 satisfies the divisibility by 7.
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coolnfundu
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Re: MS question 3
« Reply #9 on: Oct 18th, 2004, 3:55am » |
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it does not obey the second condition of divisibility by two
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Grimbal
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Re: MS question 3
« Reply #10 on: Oct 19th, 2004, 9:45am » |
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The original riddle asked about the n left-most digits to be divisible by n. So, it should read:
a % 1 = 0 (% = modulus)
ab % 2 = 0
abc % 3 = 0
abcd % 4 = 0
abcde % 5 = 0
abcdef % 6 = 0
abcdefg % 7 = 0
abcdefgh % 8 = 0
abcdefghi % 9 = 0
And 78 is divisible by 2.
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avatar
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Re: MS question 3
« Reply #11 on: May 2nd, 2008, 11:00pm » |
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What is the algo for this
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softsec
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Re: MS question 3
« Reply #12 on: Jan 27th, 2013, 2:21am » |
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brootforced?
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