# How to simpify the way of calculating $(a^b \bmod c) \bmod d$?

Right now I know the way to calculate $a^b \bmod c$ which is in the answer of the following question.

Consider the case where $c$ is much larger than $d$. So, the result of $a^b \bmod c$ will be large compared with the final result.

Is there any way to get the final result without getting directly the result of $a^b \bmod c$ which is large?

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I don't think so. By definition $(a^b \mod c) = a^b - k \cdot c$ for some $k \in \mathbb{Z}$, and in general $(a^b - k_1 \cdot c) \not\equiv (a^b - k_2 \cdot c) \mod d$ for $k_1 \neq k_2$ (unless there is some hidden relation between $c$ and $d$). So before you continue with the $\mod d$-step, you need to find the right $k$ (or the right value $a^b - k \cdot c$). –  TMM Apr 7 '12 at 21:34

If $d$ and $c$ have a factor in common, then you can use this to speed up the computation -- and in fact this is routinely done in smart cards when the RSA private key is known. But otherwise, you just have to compute $a^b \mod c$ and reduce the result modulo $d$.