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5 times 5, the units digit will always be 5. And if it times 5 again, the units stay the same.

We shall proceed to overkill this problem by induction.

Base case: $5^1=5$ , which ends in a digit of $5$ .

Induction step: we assume $5^k$ has a units digit of $5$ . This means that we can express it as $10a+5$ , where $a$ is a positive integer. We see that $\begin{aligned}5^{k+1}&=5^{k}\times 5 \\ &= (10a+5)\times 5\\ &=50a\times 25\\ &= 10(5a+2)+5\end{aligned}$

We now substitute $5a+2=b$ . Clearly, since $a$ is a positive integer, $b$ is also a positive integer. Thus, $10b+5$ also has a units digit of $5$ , and we are done. $\Box$ .

Look at the last digit and how it changes. Keep going. It doesnt change. It is still 5.

unit digit always be 5

becuase each time we multiply the results with 5 and unit digit of the result always be 5. .we can see this in the above pattern.

Thus, 5 * 5 = 25 unit digit always be 5.

its obvious that the unit will be 5 if we'll multiply 5 by any of the # in power of 5.....

so ans= 5

5^{25}=3125^{5}