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I was doing some of the previous math contests and faced a question that asked me "the number of two digit primes that are still primes when the digits are reversed".

I actually wrote down every two digit primes and then checked with the condition. Honestly it was not a hard work but it took me about five minutes only for this problem while only 30 minutes are allowed for the whole test of 15 problems. And even more, I made a small mistake of thinking "91" is a prime so got this problem wrong.

So my question is, is there any special shortcut way that I can get the prime numbers quickly and without missing? What should I do if the problem asks me about three or more digit prime numbers?
Thanks! It is #7 of Part 1 [From this test] http://www.wsmc.net/contests/2008_Contest/regtopprob.pdf

Free Spirit
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Of course you need both digits to be odd, and not to start or end with $5$.

Other than $11$, you can also eliminate cases where the digits are the same.

So you check:

$11, 13, 17, 19, 31, 37, 39, 71, 73, 79, 91, 93, 97$.

Of these, you can eliminate $39, 91, 93$ as nonprime.

From here, the answer is not hard to find: $11, 13, 17, 37, 79$ and their reverses.

For more digits, the problem won't be quite as easy since, e.g., the middle digit can be even in the three digit case. Probably a semi-smart mostly-brute-force method will be your best bet.

Benjamin Dickman
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    There is 79 and 97,too :) – Free Spirit Feb 26 '14 at 02:55
  • @FreeSpirit Right! I should've written them all out instead of saying "and their reverses." Good catch. – Benjamin Dickman Feb 26 '14 at 02:55
  • Hmm, it's a little sad that there is no exact short cut but anyway thanks so much! I see it more clearly now :) It just took me some time. Thanks – Free Spirit Feb 26 '14 at 02:56
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    @FreeSpirit Well the two digit numbers go from $11$ to $99$, which means there are a total of eighty-nine of them. A couple quick observations reduces the number of cases to thirteen. That's a pretty good shortcut, I think! – Benjamin Dickman Feb 26 '14 at 02:59
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    You could also add to your starting condition that both digits should not be multiples of 3 and the sum of the digits should not be divisible by 3. That only leaves 91 to check. – colormegone Feb 26 '14 at 02:59
  • @RecklessReckoner Yes; I considered including this condition, but thought the extra effort of mentally checking outweighed the benefits since only two numbers are eliminated in this way. This reasoning is a bit post hoc, though; your observation is nice, too. – Benjamin Dickman Feb 26 '14 at 03:02