[rookie1ja]

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.

Spoiler for 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.)

Spoiler for old wording:

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)

[shallaay]

What about 270381 would that work as well?

[muppeteer]

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.

[Aldoweb]

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.

[Slick_Rick9009]

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

[Aldoweb]

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

[KBHoleN1]

Quote

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.

[slmo]

I got 276309 by trial & error, I'm sure there are more.

Quote

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.

[onyx_omega]

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.

[eleftheria]

i agree with onyx_omega
my guess would be 692703