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Title: MS question 3 Post by kiki lee on Jun 25th, 2003, 1:36pm 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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Title: Re: MS question 3 Post by Ozzie on Jun 26th, 2003, 11:16am 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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Title: Re: MS question 3 Post by towr on Jun 26th, 2003, 11:49am 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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Title: Re: MS question 3 Post by wowbagger on Jun 26th, 2003, 11:54am 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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Title: Re: MS question 3 Post by BNC on Jun 26th, 2003, 11:54am on 06/26/03 at 11:49:54, towr wrote:
here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1045142771;start=0) |
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Title: Re: MS question 3 Post by towr on Jun 27th, 2003, 1:11am on 06/26/03 at 11:54:29, wowbagger wrote:
Not usually.. why? ;D heh, I did get it right in the other thread :P |
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Title: Re: MS question 3 Post by Claire Main on Aug 28th, 2004, 4:22pm 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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Title: Re: MS question 3 Post by Icarus on Sep 17th, 2004, 7:12pm True - but if you can find the answer for one, the other is fairly obvious, isn't it? :) |
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Title: Re: MS question 3 Post by Grimbal on Sep 18th, 2004, 11:19am 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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Title: Re: MS question 3 Post by coolnfundu on Oct 18th, 2004, 3:55am it does not obey the second condition of divisibility by two |
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Title: Re: MS question 3 Post by Grimbal on Oct 19th, 2004, 9:45am 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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Title: Re: MS question 3 Post by vijendhar on May 2nd, 2008, 11:00pm What is the algo for this |
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Title: Re: MS question 3 Post by softsec on Jan 27th, 2013, 2:21am brootforced? |
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