# Finding the last two digits of the given number.

Question:

Given $N=2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}}$
Written in the usual decimal form, find the last two digits of the number $N$.

My attempt:

We know that every exponential number repeat its last,last two,last third , $\ldots$ digits. For eg.: unit digits of $3^n$ repeats itself as $3,9,7,1,3,9,7,1...$ On applying this for last two digits of $2^n$ we get
$02,04,08,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52$ and $04,08...$ again.

This repetitive series has $21$ numbers with first one(i.e.$1$) occurring only once.
Also all $n$ in $2^n$; $5,125,625...$ when divided by $20$ give $5$ as remainder so the last two digits will be $32,16,16,16...16$ with $16$ occurring $2014$ times which gives $32+32224=32256$ which means last two digits will be $56$ but when doing the same for last digit, I get $0$ which is a contradiction.

I can't find what mistake I did.
Thanks for finding my mistake.

• You correctly determined the period of the last two decimal digits of $2^n$ to be $20$, but then you used $19$ once instead of $20$. All contributions are $32$, not just the first – joriki Jun 11 '18 at 7:46
• @joriki- Ok got it. My problem was that when we take $25$ last two digits will be 32 but I was taking 16 because I was subtracting 1 before which I had to do afterwards. – Love Invariants Jun 11 '18 at 7:52

Note that: $$02,\underbrace{04,08,16,32,64,\cdots,76,52}_{20}, \underbrace{04,08,16,32,64,\cdots,76,52}_{20},\cdots$$ To find the last two digits of $n=5,25,125,...$, you must consider: $n-1$ mod $20$: \begin{align}5-1\equiv4 \pmod{20} \Rightarrow 32\\ 25-1\equiv4 \pmod{20} \Rightarrow 32\\ 125-1\equiv4 \pmod{20} \Rightarrow 32\end{align} Alternatively, you can consider the last two digits of $(2^5)^m=32^m:$ $$\color{red}{32};24;68;76;\color{red}{32};24;68;76;\color{red}{32};24;68;76;...$$ Since $5m=5^n \Rightarrow m=5^{n-1}=1;5;25;125;... \Rightarrow m\equiv 1 \pmod 4$, then: $$\underbrace{32+32+\cdots+32}_{2015}=32\cdot 2015=64480\equiv 80 \pmod{100}.$$

We need to evaluate

$$2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}} \mod {100}$$

and

• $2^5=32 \mod {100}$
• $2^{5^2}=2^{10}2^{10}2^{5}\equiv24\cdot 24\cdot 32\equiv32 \mod {100}$
• $2^{5^3}=(2^{10}2^{10}2^{5})^5\equiv 32^5 \equiv 32\mod {100}$
• ...

therefore

$$2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}} \equiv2015\cdot 32\equiv 80\mod {100}$$

• Thx, but I want to find my mistake. – Love Invariants Jun 11 '18 at 7:46
• According to the period you have found we have that $2^5\equiv 32$ and $2^{25}\equiv 32$ and so on. – user Jun 11 '18 at 7:53
• I got what I mistake I did. That was only because I was subtracting 1 from all $n$ because 2 was not repeating so I had to take $4^{th}$ in the recurring series not the $4^{th}$ in full series. – Love Invariants Jun 11 '18 at 7:56
• @LoveInvariants you have 02 only for the case $2^1$ and the period is 20 thus $2^{5+20k}\equiv 32 \mod {100}$ – user Jun 11 '18 at 7:58
• Yeah thanks ... – Love Invariants Jun 11 '18 at 7:59