rookie1ja 13 Posted March 30, 2007 Report Share Posted March 30, 2007 Cipher - Back to the Number Puzzles Find the number if: 1. The cipher is made of 6 different numerals. 2. Even and odd digits alternate, including zero (in this case as an even number). 3. The difference between two adjacent numerals is always greater than one (in absolute value). 4. The first two numerals (as one number) as well as the two middle numerals (as one number) are a multiple of the last two numerals (as one number). What is the cipher? There is more than 1 solution. This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Cipher - solution The possible 2 last numerals are as follows: 03, 05, 07, 09, 14, 16, 18, 25, 27, 29 and 30. At least two multiples less than 100 (this condition is already accomplished), which consist of even and odd numeral (respecting all other conditions) are for 03, 07, 09 and 18 as follows: 03 – 27, 63, 69, 81 07 – 49, 63 09 – 27, 63, 81 18 – 36, 72, 90 There are 5 numbers that can be made of these pairs of numerals to create the cipher: 692703, 816903, 496307, 816309 and 903618. (If we assume, that also in the number 903618 is accomplished the requirement to alternate even and odd numbers, despite the opposite order.) Find the cipher if: 1. The cipher is made of 6 different numerals. 2. Even and odd figures alternate, including zero (in this case as an even number). 3. The difference of adjacent numerals is always bigger than one. 4. The first two numerals (as one number) as well as the two middle numerals (as one number) are a multiple of the last two numerals (as one number). What is the cipher? (more than 1 solution) Link to post Share on other sites

Guest Posted June 29, 2007 Report Share Posted June 29, 2007 What about 270381 would that work as well? Link to post Share on other sites

Guest Posted July 3, 2007 Report Share Posted July 3, 2007 last two digits cannot end in a 0 (all multiples will have a zero in them, thus 10, 20, 30, ... 80, 90 not the last two digits) [removes 30 from your list] last two digits must be less than 33 (to satisfy 1 and 4) last two digits cannot be adjacent, nor both even (from 2) [removes 14] last two digits cannot have multiples less than 100 whose digits are adjacent [removes 27, 29 from your list] last two digits cannot have a 5 in it (05 and 25 removed - no pair of multiples lack both 5 and 0) last two digits cannot have multiples comprised only of even digits [removes 16] 03, 07, 09, 18 remain 03 can use the following multiples: 15, 18, 27, 69, 81 07 can use the following multiples: 49, 63, 91 09 can use the following multiples: 18, 27, 36, 63, 72, 81 18 can use the following multiples: 36, 72, 90 example: pick a number ending in "07" and use 49 and 63 gives us 496307 or 634907 However, because of the different rules, not every combination is allowed 03 only has 9 possible combinations 07 only has 2 possible combinations 09 only has 17 possible combinations 18 only has 6 possible combinations Thus, I believer there to be only 29 different cyphers possible, according to the stated boundary conditions. Link to post Share on other sites

Guest Posted July 20, 2007 Report Share Posted July 20, 2007 Have you forgot the easiest of all What about 01 as two last numerals. Anything else can be on the other four places obeying only to the first condition. Link to post Share on other sites

Guest Posted July 26, 2007 Report Share Posted July 26, 2007 01 wouldn't work 'cause 0 and 1 have a difference of 1. Link to post Share on other sites

Guest Posted July 31, 2007 Report Share Posted July 31, 2007 No man Read 3. The difference of adjacent numerals is always bigger than one. Here the puzzle requests the difference of the cipher 1 and 6 to be more than 1. Prove me wrong Link to post Share on other sites

Guest Posted August 28, 2007 Report Share Posted August 28, 2007 No man Read 3. The difference of adjacent numerals is always bigger than one. Here the puzzle requests the difference of the cipher 1 and 6 to be more than 1. Prove me wrong Ok. 3. The difference of ADJACENT numerals is always bigger than one. If the last two numerals are "01", then adjacent numerals have a difference (1-0=1). Last time I checked, 1 is not bigger than 1. Perhaps you should learn to read, man. Link to post Share on other sites

Guest Posted October 9, 2007 Report Share Posted October 9, 2007 I got 276309 by trial & error, I'm sure there are more. No man Read 3. The difference of adjacent numerals is always bigger than one. Here the puzzle requests the difference of the cipher 1 and 6 to be more than 1. Prove me wrong "adjacent" means next to, not the 1st and 6th digit of the final number. Link to post Share on other sites

Guest Posted November 10, 2007 Report Share Posted November 10, 2007 Shallaay, 270381 would be the only working solution if the problem had stated "factors" instead of "multiples." I did the same thing, and that's what I got. If that were the puzzle, there would only be one answer, in which case, I think it would be a better puzzle. Link to post Share on other sites

Guest Posted November 16, 2007 Report Share Posted November 16, 2007 i agree with onyx_omega my guess would be 692703 Link to post Share on other sites

Guest Posted March 13, 2008 Report Share Posted March 13, 2008 Cipher - Back to the Number Puzzles Find the cipher if: 1. The cipher is made of 6 different numerals. 2. Even and odd figures alternate, including zero (in this case as an even number). 3. The difference of adjacent numerals is always bigger than one. 4. The first two numerals (as one number) as well as the two middle numerals (as one number) are a multiple of the last two numerals (as one number). What is the cipher? (more than 1 solution) Cipher - solution The possible 2 last numerals are as follows: 03, 05, 07, 09, 14, 16, 18, 25, 27, 29 and 30. At least two multiples less than 100 (this condition is already accomplished), which consist of even and odd numeral (respecting all other conditions) are for 03, 07, 09 and 18 as follows: 03 – 27, 63, 69, 81 07 – 49, 63 09 – 27, 63, 81 18 – 36, 72, 90 There are 5 numbers that can be made of these pairs of numerals to create the cipher: 692703, 816903, 496307, 816309 and 903618. (If we assume, that also in the number 903618 is accomplished the requirement to alternate even and odd numbers, despite the opposite order.) I'm not seeing how the first two and second two numerals are a multiple of the last two numerals in 496307. Can you help me out? Link to post Share on other sites

Guest Posted March 13, 2008 Report Share Posted March 13, 2008 I'm not seeing how the first two and second two numerals are a multiple of the last two numerals in 496307. Can you help me out? nevermind I see it now. I thought you meant the sum of the two numerals. which surprisingly works for all the other solutions. Link to post Share on other sites

Guest Posted March 26, 2008 Report Share Posted March 26, 2008 (edited) What about 270903? EDIT: ...wait... That doesn't meet the 1st requirement that they be "different" numerals. so, 274503 should meet the requirements. 2nd EDIT: Crap. the 4 and 5 are consecutive. hmmm.... Edited March 26, 2008 by gmarsha11 Link to post Share on other sites

Guest Posted April 12, 2008 Report Share Posted April 12, 2008 (edited) 692703 is what I've come up with. Not that my entry does it, but the OP could've made it more interesting if he asked for the 'smallest' possible number meeting the criteria. Edited April 12, 2008 by demented Link to post Share on other sites

Guest Posted June 14, 2008 Report Share Posted June 14, 2008 I got 276309 by trial & error, I'm sure there are more. "adjacent" means next to, not the 1st and 6th digit of the final number. 276309 has 7 & 6 next to each other, and also the questioner doesn't state whether the numerals 9 & 0 should be regarded as adjacent (as they are on a bicycle padlock for example) Link to post Share on other sites

Guest Posted June 14, 2008 Report Share Posted June 14, 2008 Well I think there are only two possible solutions, if 9 and 0 are to be regarded as adjacent and therefore excluded. 692703 and 496307 these others use 9 & 0 together: 903618; 816903 and 816309 Link to post Share on other sites

Guest Posted August 5, 2010 Report Share Posted August 5, 2010 (edited) Am I correct in assuming there are exactly 65 missed a couple adjacent possible soutions? or 2 if 0 and 9 are considered adjacent Edited Edited August 5, 2010 by Polynym Link to post Share on other sites

Guest Posted September 3, 2010 Report Share Posted September 3, 2010 (edited) thank you to everyone for using a spoiler..much obliged Edited September 3, 2010 by Brain Tickler Link to post Share on other sites

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