Came across some notes on curious properties of the number 17 in a blog, called Mathnexus. Then today one of my students announced that she was 17 years old... So I shared them with her and the class......and now with you:

It is the only known prime that is equal to the sum of digits of its cube (17^{3} = 4913, and 4 + 9 + 1 + 3 = 17)

It is the only prime that is the average of two consecutive Fibonacci numbers. ... (Ok, that would be 13 and 21... now the only way Fibonacci numbers can have an integer value is if they are both odd *[there are no consecutive even Fibonacci numbers}*, and all the even Fibonacci numbers are twice the average of the two previous values... so there can be no Fibonacci number after 34 which is twice a prime) It is interesting that there are lots of odd prime Fibonacci numbers, 2, 3, 5, 13, 89, 233, 1597, 28657, for example , Sloane's A005478, and each of them has a prime index (except three, which is f_{4}).

It is the least integer such that the sum of its digits in every base B = 2, 3, 4, 5, 6, 7, 8 is prime. (In base two ,17 is represented as 10001 and 1+0+0+0+1=2, a prime number; in base three it becomes 122, 1+2+2=5; in base four it is 41, in base five it is 32; in base six it is 25; in base seven it is 23; and in base eight it is 21... but in base nine it becomes 18, and 1+8 = 9 is not prime.) which makes me wonder... is there a prime number for which the sum of the digits in all bases two through ten is also prime?

There are exactly 17 ways to express 17 as the sum of one or more primes.(can you list them all?)

And just one more, there is no odd Fibonacci number that is divisible by 17. (Ok, how special IS that?... are there other (odd) numbers that do not divide evenly into any of the odd Fibonacci numbers?..YES, the smallest is nine. A good strategy for attacking this kind of problem is given in this blog by Tanya Khovanova. Tanya goes on to state that none of the odd Fibonacci numbers are divisible by 19, 23, or 27 also, so maybe this really isn't such an unusual event at all.)