# Prime number puzzle

1. Dec 9, 2008

### Loren Booda

What three digit prime number has all prime digits and forms primes with its first two and last two digits?

2. Dec 10, 2008

### kenewbie

523

k

3. Dec 10, 2008

### d_leet

It's
373

4. Dec 10, 2008

Well it can't be 523 - 52 isn't prime.

d_leet's answer is the only one.

5. Dec 10, 2008

### noisysignal

i guess its 373 (i had to add the "i guess its" because PF told me my post has to be atleast 4 charachters long :P)

6. Dec 12, 2008

### kenewbie

Ah. I misread the OP. I thought the *sums* d1 + d2 and d2 + d3 had to be prime.

ie 5 +2 = 7 and 2 + 3 = 5

At least my answer can be used to form a new puzzle :p

k

7. Dec 12, 2008

### dodo

A quick&dirty computer program shows:
11 13 113
13 31 131
13 37 137
17 73 173
17 79 179
19 97 197
31 11 311
31 13 313
31 17 317
37 73 373
37 79 379
41 19 419
43 31 431
47 79 479
61 13 613
61 17 617
61 19 619
67 73 673
71 19 719
79 97 797
97 71 971

8. Dec 12, 2008

Remember, each digit has to be prime too, so the only one that works is 373. :)

9. Dec 14, 2008

### robert Ihnot

Noisy signal has seen 373, and adriank has seen after the programming of Dodo, that it is the only solution.

This result can be achieved by logic: We look at the four one digit primes: 2, 3,5,7. We see that 5 can not be in the second or third place, and neither can 2. Any number containing digits 5,3,7 or 2,3,7 is divisible by 3. So only 3 and 7 can be used. 7 can not be in two successive places, and neither can 3. 737 is divisible by 11, so only 373 is left!

Last edited: Dec 14, 2008
10. Dec 14, 2008

Testing each number from 100 to 999 is logic too.

11. Dec 14, 2008

### Loren Booda

Perhaps robert's solution is more elegant?

12. Dec 14, 2008

Certainly; if I was without a calculator or so that's probably how I would do it. (That was probably the intention of the problem anyway.)

13. Dec 15, 2008

### robert Ihnot

adriank: Testing each number from 100 to 999 is logic too.

I kept thinking I should say, "By logic alone," but I thought that might sound too egotisical. (But? Oh, well!)

14. Dec 15, 2008

### dodo

I'm still beating myself over not even imagining that the problem was solvable by logic.

15. Dec 17, 2008

### robert Ihnot

Well, I might add here, in case there was a problem. If the sum of the digits is divisible by 3 the number is divisible by 3. That takes care of any three terms containing 237, or 537. Now, a number is divisible by 11 if the alternating sum of its digits is divisible by 11. Here we have 7-3+7=11.

Of course, years ago people knew those things, but since calculators, well, I am not so sure! In checking 373, we can eliminate 2,3,5,7,11 by simple rules. (For 7 the rule is to double the last term and subtract it from the rest.) Thus 37-6 is not divisible by 7.
(That rule is really neat because 91x7 =637, but to check for 7 it is only necessary to subtract 14 from 63, leaving 49.)

Possible factors of 373 such as 13,17,19 then could be checked by long division or a calculator.

Last edited: Dec 17, 2008