Find the last digits without the calculator

Method $1$ : Chinese remainder theorem.

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Method $2$ : pattern spotting

We know that for $n>1$ , the last two digits are always $25$ .

Define a sequence $a_n$ by $5^n = 100a_n+25$ for $n>1$ .

We have $a_2=0$ and $5^{n+1}=100a_{n+1}+25=500a_n+125$

so $a_{n+1}=5a_n+1$ . We're only interested in the last three digits of this number, so we can ignore the others. The sequence goes $0,1,6,31,156,781,906,531,656,281,406,31,156,781,906,\cdots$

and we see it locks into a repeating cycle of $8$ numbers. Using this pattern, we find the last three digits of $a_{55}$ are $781$ so that the last $5$ digits of $5^{55}$ are $\boxed{78125}$ .

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Method $3$ : better pattern spotting

The last three digits of $25^n$ are $625$ for $n>1$ . Define $25^n=1000b_n+625$ ; we then get $b_2=0$ , $b_{n+1}= 25b_n+15$ , and we're only interested in the last two digits of $b_n$ .

Even better, $b_n$ is always a multiple of $5$ , so $20b_n$ is always a multiple of $100$ ; so we just need to keep track of the last two digits of $b'_n$ where $b'_2=0$ and $b'_{n+1}=5b'_n+15$ .

The sequence we get is $0,15,90,65,40,15,90\cdots$ , which gets into a cycle of period $4$ . So the last two digits of $b'_{27}$ are $15$ ; the last five digits of $5^{54}$ are $15625$ , and we get the same answer.

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Method $4$ : do you know your powers of $625$ ?

There's an even stronger pattern in the last five digits of powers of $625$ , though they're harder to work out without a calculator; this sequence is $625,90625,40625,90625,\cdots$

From this, $5^{52}$ ends in $40625$ , and we can multiply by $125$ to get the answer.

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Other methods:

Decimal expansions of $2^{-n}$ have the same digits as powers of $5$ ( $0.5,0.25,0.125,0.625,\cdots$ ); is there a way to use this?

By Chinese remainder theorem , we have $5^{55} \bmod 5^5 = 0$ ; and for $5^{55} \bmod 2^5$ , we note that $\gcd(5, 2^5) = 1$ , we can apply the Euler's theorem and note that Euler's totient function $\phi(2^5) = 2^5 \times \frac 12 = 16$ . Then

$\begin{aligned} 5^{55} & \equiv 5^{55 \bmod \phi(32)} \equiv 5^{55 \bmod 16} \equiv 5^7 \text{ (mod 32)} \\ & \equiv 5(125)(125) \equiv 5(-3)(-3) \equiv 45 \equiv 13 \text{ (mod 32)} \end{aligned}$

$\begin{aligned} \implies 5^{55} & \equiv 32 n + 13 & \small \blue{\text{where }n \text{ is an integer.}} \\ \implies 32n + 13 & \equiv 0 \text{ (mod 3125)} & \small \blue{\text{Note that }5^5 = 3125} \\ \implies n & \equiv 2441 \\ \implies 5^55 & \equiv 32(2441) + 13 \equiv \boxed78125{} \text{ (mod 100000)} \end{aligned}$

On other ways of solving this problem , of course the easiest "way" to solve this is to hit "5^55 mod 100000" in WolframAlpha . Of course, you would said it is numerical method. But in fact is solving $32n+13 \equiv 0 \text{ (mod 3125)}$ , I have resorted to Python to solve $n=2441$ , the other solutions are $n=5566, 8691$ . Also I think the pattern spotting method proposed by @Chris Lewis would also need a computer to generate the patterns. I just used the Coding environment of Brilliant to get the following pattern. It has a period of $8$ .

$\text{I used a calculator 😈😜. AND REKTED YOU ALL!}$

$\text{P. S. It was}$ $78125$ $\text{. I used a calculator app and had to GODDAMN scroll for}$ $\text{almost eternity[insert rage here]}$ $\text{(aka. 3 seconds).}$ $\text{P. P. S. All this writing is also taking almost eternity [insert more rage here](aka. 10 GODDAMN minutes)}$