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If i have a really large number like \begin{gather} 3^{762259784987} \end{gather} or \begin{gather} 7^{836793257}\end{gather} Is there a way to find the last three digits without using modular arithmetic or is that the only method. If their is could someone please outline a method because i would be very curious. I don't want to use modular arithmetic, despite it probably being the most obvious method, because we haven't covered the topic yet but I still received a similar problem which leads me to believe that their could be another method that relies more on problem solving skills. But I could be very wrong and maybe the only method involves modular arithmetic. I worked out how to find the last digit because of the cyclical nature but I haven't had any luck with figuring out all of the last three digits.

(Note: Someone rightfully pointed out that the title and the question asked for two different things, my fault and sorry for any confusion but I was originally thinking about the last three)

XAVIER2004
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  • So a "workaround" would be simply to do several $3^n$ and $7^m$ and find a pattern. This is ultimately the same as modular arithmetic but without that machinery. – Cameron Williams Feb 16 '21 at 13:20
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    See for example here. After all, modular arithmetic is not a heavy machinery. Try it. – Dietrich Burde Feb 16 '21 at 13:21
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    Anything you do will be more or less equivalent to modular arithmetic. In this case, you could work with $5^2$ and $2^2$ instead of $10^2$ which greatly simplifies the computation. – lulu Feb 16 '21 at 13:21
  • @lulu $3^{87}$ has the same last three digits as $3^{762259784987}$. Is there a link with the fact that the "period" of $3^n\bmod 1000$ is $100$ and $762259784987\bmod 100 = 87$? – Raffaele Feb 16 '21 at 13:51
  • It is very easy to find $c$ in $N^n = 10M + c$ by the periodicity modulo $4$ of the successive powers of the digits. Something similar could happen with $N^n = 1000M + abc$. With the help of a computer this can be ascertained in advance and after this the calculation of $abc$ will be easy. – Piquito Feb 16 '21 at 13:57
  • @Raffaele Sure. $3^{100}\equiv 1 \pmod {10^3}$. Of course, these periods are kind of long for somebody trying to do this by pencil and paper arithmetic. – lulu Feb 16 '21 at 13:59
  • The last three digits of a number are that number modulo $1000$. Just this statement tells you that it will be difficult to do without modular arithmetic ;-) –  Feb 16 '21 at 14:18

1 Answers1

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Update: Last three digits of $3$ requires lots of work but not impossible as has already been in the comments. With the last three digits of $7$, we are a little bit luckier, though:

1 7 <-
2 49
3 343
4 401
5 807
6 649
7 543
8 801
9 607
10 249
11 743
12 201
13 407
14 849
15 943
16 601
17 207
18 449
19 143
20 1
21 7 <-

So, the last three digits are repeated per $20$ steps.

Looking at the two last digits of powers of $3$ until the repetition, we get the following list:

1 3 <-
2 9
3 27
4 81
5 43
6 29
7 87
8 61
9 83
10 49
11 47
12 41
13 23
14 69
15 7
16 21
17 63
18 89
19 67
20 1
21 3 <-

So, the last two digits are repeated per $20$ steps.

Looking at the two last digits of powers of $7$ until the repetition, we get the following list:

1 7 <-
2 49
3 43
4 1
5 7 <-

So, the last two digits are repeated per $4$ steps.

VIVID
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