BrainDen.com - Brain Teasers

# Cipher

## Recommended Posts

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)

##### Share on other sites

• 2 months later...

What about 270381 would that work as well?

##### Share on other sites

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.

##### Share on other sites

• 3 weeks later...

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.

##### Share on other sites

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

##### Share on other sites

No man

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

##### Share on other sites

• 4 weeks later...
No man

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.

##### Share on other sites

• 1 month later...

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

No man

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.

##### Share on other sites

• 1 month later...

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.

##### Share on other sites

i agree with onyx_omega

my guess would be 692703

##### Share on other sites

• 3 months later...
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?

##### Share on other sites

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.

##### Share on other sites

• 2 weeks later...

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 by gmarsha11
##### Share on other sites

• 3 weeks later...

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 by demented
##### Share on other sites

• 2 months later...
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)

##### Share on other sites

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

##### Share on other sites

• 2 years later...

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 by Polynym
##### Share on other sites

• 4 weeks later...
thank you to everyone for using a spoiler..much obliged Edited by Brain Tickler
##### Share on other sites

This topic is now closed to further replies.