Can this be solved with any HP CAS?

[Nigel (UK)]

On the Prime, two ways:

• irem(2^1000,13) where irem is the integer remainder function;
• powmod(2, 1000, 13) which is more efficient, because it finds the remainder without first working out the power of two.
  

It doesn't take long with pencil and paper either!

Nigel (UK)