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 $n>1$
 A: The answer of Alonso del Arte is good, but you can do this almost without calculations :
You are interested by $n>1$. So you can say, for $n\ge 0$ it's equivalent to :
$$\phi(n)=1^{2^n}+16^{2^n}+81^{2^n}+256^{2^n}$$
$100=2^2.5^2$, so you can decompose the problem modulo 4 and modulo 25.


*

*Modulo $4$, $1\equiv81\equiv1[4]$ and $16\equiv256\equiv0[4]$. So
$$\phi(n)\equiv2[4]$$ 

*Modulo 25, $16^2\equiv 6[25]$, $6^2\equiv11[25]$, $11^2\equiv21[25]$, $21^2\equiv16[25]$ (you can all compute mentally, and remember, $21\equiv -4[25]$).
The easy part comes from the fact that $81\equiv256\equiv6[25]$. So each term have the same period of 4 values (16,6,11,21). You can check that each time, the sum will be 4 (for the first one, $1+16+6+6\equiv 4[25]$, for example)
$$\phi(n)\equiv 4[25] $$


Now, there is only one number that verifies both equality modulo 100, it's 54, thanks to the Chinese remainder theorem.
A: 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.
A: You can prove it by demonstrating that the last two base $10$ digits of $2^{2^n}(2^{2^n} + 1)$ and $3^{2^n}$ become periodic past $n = 1$ and that the length of that period is $4$. (By the way, it was bugging me that $2$ and $4$ are not coprime, so I tried to see if there was a simpler expression for $4^{2^n} + 2^{2^n}$ but I couldn't find it).
The period for $4^{2^n} + 2^{2^n}$ is $\{72, 92, 32, 12\}$. The period for $3^{2^n}$ is $\{81, 61, 21, 41\}$. Adding those two sequences together we get $\{153, 153, 53, 53\}$.
