How old are you?

Three cubed. That feels pretty good.

Yay This Is Fun! , I'm 23 year's of age and not counting.

1234321 wrote:Three cubed. That feels pretty good.

Precisely.

BTW, 1234321, I see you already know this in the What inspired your username? thread, but I was surprised to find out that your user name is the square of another palindromic number! What's more, the root of your name factors into two palindromic primes.
I guess I could call you "(11x101)x(101x11)".

Some more that you might already know, but I find interesting: Your age is the smallest square of a square after 1, the smallest prime power of a prime power of a prime (2^2^2), the smallest square whose reverse is prime, and the only integer that equals m^n and n^m, for some unequal integers m and n, (2^4 & 4^2). It has this property because 2^2 = 2 × 2. Of course, because it is the 4th power of 2, your age has been used in measuring objects in several cultures, (16 oz./lbs., 16 liangs/jin, etc.).

I like the 2^2^2 thing. Numbers are so interesting.

21

29

lol im 17, way to be a math wizz, lol yes i am foolish

Even though I like numbers a lot, I'm a really slow learner when it comes to math. I'll stop after this unless I get requests.

21:
The sum of the first six integers (1+2+3+4+5+6=21)

29:
The number of times The Beatles pronounce "yeah" in their 1963 hit "She Loves You" (... yeah, yeah, yeah).

17:
The only known prime that is equal to the sum of digits of its cube (17^3 = 4913 and 4+9+1+3=17)
The only prime of the form p^q + q^p, where p and q are prime (17 = 2^3 + 3^2).
The smallest multidigit prime which requires exactly a prime number of letters to spell in both English and Latin ("seventeen" has 19 letters, and "septemdecim" has 11 letters)
If you were a wizard, you'd be considered "of age." Just don't get splinched!

S3 wrote:The only prime of the form pq + qp, where p and q are prime (17 = 23 + 32).

How does this work?

oops. I forgot the carats. good catch.
17 = p^q + q^p = 2^3 + 3^2

I don't have the proof in front of me, but the statement above implies that while some primes may be expressed in the same form in terms of other primes, (p^q + r^s), the primes p, q, r, and s will always be 3 or 4 distinct primes, unlike (2^3 + 3^2).

Here's my best guess at why ~

Given: p, q, and x are prime, and x = p^q + q^p

An odd number times an odd number is always odd.

x = (odd#)^(odd#) + (odd#)^(odd#) = odd# + odd#
= an even number
So x is never a prime when neither p nor q equals 2.

If p = 2 but q ≠ 3 then...
2^q = (some multiple of 3) - 1.
q^2 is never a multiple of 3 so...
q^2 = (some multiple of 3) - 2.

x = 2^q + q^2 = (some multiple of three) +(some multiple of three) - 1 - 2 = (some multiple of 3) + (some multiple of 3) - 3
= a multiple of three
So x is never a prime when p = 2 but q ≠ 3.

Therefore p=3, q=2, and x=17

If that proof is correct then I'm awesome...