numbers by the book

[phil1882]

Five men, one of which is Greg each selected 5 unique 3 digit numbers from a hat. given the clues, can you tell what the numbers are and the men who selected them?

1. Each digit for the number chosen by each man is unique. (that is, if the number begins with 1, 1 wont appear anywhere else in the number.)

2. Each digit for each place between men is unique. (that is, if a number begins with 1, no other number begins with 1.)

3. Only the digits 1-5 are used.

4. Andy has the smallest number, and it's evenly divisible by 5.

5. Both Bob's and Fred's numbers are evenly divisible by 3.

6. Bob's number is smaller than Fred's.

7. Tim's number is evenly divisible by 2, and a prime.

8. Tim's number is greater than Bob's.

9. Fred's middle most number is prime.

10. Bob's middle and last number are odd.

11. the middle digit of Tim's number is twice as big as Andy's.

 

good luck and enjoy!

[wolfgang]

Sorry...may I ask a question?.

Is it true for the digits in the middle ot at the end  of the choosen number to be unique also?

[wolfgang]

If the first digit is unique...so

There are alot of possible solutions , one of these is:-

Andy........125

Bob.........213

Tim..........342

Fred........432

greg........513

[phil1882]

this solution doesn't work, as tim and fred both end in 2.

the numbers 1-5 should appear once in each digit place between men, as hint #2 tries to explain.

 

[harey]

I have got 3 solutions, but I do not really understand the point (9)

Andy: 1AB (4)  - the smallest number
      1A5 (4)  - divisible by 5
      125 (11) - A=1 or 2, 1 excluded (2)

Tim:
1st digit: at least 3 (8)
2nd digit: 4 (11)
3rd digit: 2 (7), Bob has at least 2
342/2=171: not prime
542/2=271: prime

Bob:
1st digit: 2 3 4 (1 excluded, (Andy); 5 excluded (6))
2nd digit: 1 3 5 (10)
3rd digit: 1 3   (10) (5 excluded (Andy))
w/o repetition (1) and divisible by 3 (5) leaves:
213
231
351
453

Fred:
1st digit: 3 4 (1 excluded, (Andy); 2 excluded (6); 5 excluded (Tim)
2nd digit: 3 5 (10) (1 is not considered as prime)
3rd digit: 1 2 3 4 (5 excluded (Andy))
w/o repetition (1) and divisible by 3 (5) leaves:
351
354
453

This excludes 453 for Bob (6), leaving for Bob:
213
231

a) Let's assume Fred has 354:
Greg:
1st digit: 4:    7-fred (1 andy, 2 bob,  3 fred, 5 tim)
2nd digit: 1, 3: 4-bob  (2 andy, 4 tim,  5 bob)
3d digit:  1, 3: 4-bob  (2 tim,  4,greg  5 andy)
-> 413 compatible with Bob (1st solution)
-> 431 compatible with bob (2nd solution)

a) Let's assume Fred has NOT 354:
fred: 351 or 453
bob:  213 or 231
-> greg's 4th digit=4
-> fred 453 (someone has to have the first digit 4)
-> bob  231 (213 excluced)
-> andy 125
-> tim  542
-> greg 314 (3rd solution)

[phil1882]

alright sorry, i tried to make it such that there was only 1 solution but looks like i failed. thanks for finding all 3.

[harey]

What about a supplementary rule excluding Fred has 354?