<s>[INST] How can I prove the last two digits of $1+2^{2^{n}}+3^{2^n}+4^{2^n}$ always are $54$ How can I prove the last two digits of  $$1+2^{2^{n}}+3^{2^n}+4^{2^n}$$ are $54$ when $n$ is a positive integer number if $n>1$
 [/INST] These things go in cycles with periods of length $4$.
The last two digits of $4^{2^n}$ are $16$, $56$, $36$ or $96$.
For $n > 4$, the last two digits of $3^{2^n}$ are $41$, $81$, $61$ or $21$.
For $n > 4$, the last two digits of $2^{2^n}$ are $96$, $16$, $56$ or $36$.
Then notice that $16 + 41 + 96 + 1 = 154$, and therefore if $n > 4$ and $n \equiv 1 \pmod 4$ then $4^{2^n} + 3^{2^n} + 2^{2^n} + 1 \equiv 54 \pmod{100}$. You should be able to work out $n \equiv 2 \pmod 4$ and the other cases yourself.
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