Five 5s

Okay, so this question is a fairly simple application of modular arithmetic.

We know that $5^{3} = 125$ , and thus $5^{3} \equiv 125 \pmod{1000}$ .

Thus, $5^{4} \equiv 5*5^{3} \equiv 5*125 \equiv 625 \pmod{1000}$ .

Now, see that: $5^{5} \equiv 5*5^{4} \equiv 5*625 \equiv 3125 \equiv 125 \pmod{1000}$ .

From here, we can use finite induction to prove that for all natural numbers $x$ , such that $x > 2$ , $5^{x} \equiv 125 \pmod{1000}$ if $x$ is odd, and $5^{x} \equiv 625 \pmod{1000}$ if $x$ is even.

Since all natural powers of $5$ are odd numbers, then $5^{5^{5^{5}}}$ is an odd number, and so $5^{5^{5^{5^{5}}}}$ is an "odd power" of five, thus giving us the last three digits to us as being $125$ .

The given number, which we'll call $N$ , can be written in the form $5^{2k + 1}$ for some positive integer $k$ . So $N = 5^{2k + 1} = 5(5^{2k}) = 5(25^k)$ .

Now, the last three digits of $25^k$ are always 625, so we know that $5(25^k) \hspace{.15cm} \equiv 5(625) \hspace{.1cm} \equiv \hspace{.1cm} 125 \pmod{1000}$

So, the answer is 125.

The last digit of any number $5^{x}$ is always $25$ . And, another little observation may state that it is always $125$ if the power is odd, or it is $625$ when even, always. So, since it is odd, it is $125$

Now, that is a high 5 :)

5^5=3125. Thus, last three digits are 125.

Apart from 5^1 and 5^2, for a positive integer n, 5^n has 125 or 625 as its last 3 digits. 3 attempts to the question are more than enough...

5^5^5^5^5=3125^5^5^5=abc, def, ghi, jkl, mno, 125^5^5= Always ends 125 to the last 3 digits.

5^1 = 5, 5^2 = 25, 5^3 = 125 5^ 4 = 625, 5^ 5 = 5 * (625) = (10 * 625)/2 = 3125 which ends with 125, thus we cycle back to 625 for 5^6 and so on (odd power -> ends with 125, even power -> ends with 625).

We also know this patterns starts with 5^5, and that an odd number times an odd number is still odd, therefore 5's exponent 5^5^5^5, is odd and so 5^5^5^5^5 ends with 125

5 to the power of 5 is 125 And the first digits of 125 to the power of 5 is 125 and since the larger digits cant effect the smaller digits it stays 125

The fact $5^{n}$ for $n>1$ & $n$ is a natural number that the 2 last digit of $5^{n}$ (from the right) is 25

Since 5 is a odd number so the last 3 digit of $\LARGE \color{#3D99F6}5^{\color{#20A900}5^{\color{#D61F06}5^{\color{#624F41}5^\color{magenta}5}}}$ is 125

My solution was very simple solved $5^{2n+1}\mod 1000 = 125$ and i knew that $5^{5^{5^{5}}}$ was odd because a exp base 5 always end with 5

so the awnser its 125

5^5 = 3125 Mod 1000

3125 is Congruent to 125 Mod 1000

Now 125^5 = 125 Mod 1000

The rest of the expressions will be 125^5.

So the remainder will always be 125.

$\frac{\LARGE5^{5^{5^{5^5}}}}{\LARGE1000} = \frac{\LARGE5^{(5^{5^{5^5}}-3)}}{\LARGE8}$ Cancel 125 from numerator and denominator.

Now check the remainder by 8 on remaining number.

$\phi(8) = 4$

$\therefore 5^{4k} \equiv 1 \pmod{8}$

$5^{5^{5^5}}-3 \equiv 2 \pmod{4}$ $\therefore 5^{(5^{5^{5^5}}-3)} \equiv 5^2 \equiv 1 \pmod{8}$

multiply $125$ on the expression $5^{(5^{5^{5^5}}-3)}*125 \equiv 125 \pmod{8*125}$ $\therefore 5^{5^{5^{5^5}}} \equiv 125 \pmod{1000}$

Hence answer is $125$