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# Cipher

17 replies to this topic

### #1 rookie1ja

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Posted 30 March 2007 - 07:23 PM

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.

Spoiler for Solution

Spoiler for old wording

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### #2 shallaay

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Posted 29 June 2007 - 07:13 PM

What about 270381 would that work as well?
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### #3 muppeteer

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Posted 03 July 2007 - 04:31 PM

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.
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### #4 Aldoweb

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Posted 20 July 2007 - 12:37 PM

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.
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### #5 Slick_Rick9009

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Posted 26 July 2007 - 08:07 PM

01 wouldn't work 'cause 0 and 1 have a difference of 1.
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### #6 Aldoweb

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Posted 31 July 2007 - 11:41 PM

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
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### #7 KBHoleN1

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Posted 28 August 2007 - 03:49 PM

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.
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### #8 slmo

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Posted 09 October 2007 - 09:49 PM

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.
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### #9 onyx_omega

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Posted 10 November 2007 - 08:52 PM

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.
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### #10 eleftheria

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Posted 16 November 2007 - 03:42 PM

i agree with onyx_omega
my guess would be 692703
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