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- Jan 31st 2013, 02:19 PMdave52find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)
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- Jan 31st 2013, 06:04 PMSorobanRe: find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)
Hello, dave52!

Quote:

$\displaystyle \text{(a) Find }\,2^{3^{100}}\text{ (mod 5)}$

$\displaystyle \text{(b) Find the last digit of }\,2^{3^{100}}$

Here is the first one . . .

Consider the first few powers-of-2:

. . $\displaystyle \begin{array}{ccc} 2^1 &\equiv& 2\text{ (mod 5)} \\ 2^2 & \equiv& 4\text{ (mod 5)} \\ 2^3 &\equiv& 3\text{ (mod 5)} \\ 2^4 &\equiv& 1\text{ (mod 5)} \\ 2^5 &\equiv& 2 \text{ (mod 5)} \\ \vdots && \vdots \end{array}$

We see that the powers-of-2 have a cycle of 4 steps.

Now what about $\displaystyle 3^{100}\,?$

We find that: .$\displaystyle 3^{10} \:=\:59,\!049 \:\equiv\:1\text{ (mod 4)}$

. . . . . . And:.$\displaystyle \left(3^{10}\right)^{10} \:\equiv\:1^{10} \:\equiv\:1\text{ (mod 4)}$

. . . . Hence: .$\displaystyle 3^{100} \:=\:4k+1$, for some integer $\displaystyle k.$

So we have: .$\displaystyle 2^{3^{100}} \;=\;2^{4k+1} \;=\;2^{4k}\cdot2^1 \;=\;(2^4)^k\cdot 2 $

Therefore: .$\displaystyle (2^4)^k\cdot2 \;\equiv\;1^k\cdot2\text{ (mod 5)} \;\equiv\;2\text{ (mod 5)}$

- Jan 31st 2013, 08:38 PMjohngRe: find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)
There are two interpretations of 2^3^100

1. As above and with most calculators $\displaystyle 2^{3^{100}}$; mod 5, this is 2 as above. So the last digit d is 2. Note to calculate 3^100 mod 4, it's probably easier to notice 3 is -1 mod 4, so 3^100 is (-1)^100 = 1 mod 4.

2. $\displaystyle 2^{3\cdot 100}$ -- this is treating ^ as associating left to right; i.e. a^b^c is (a^b)^c as opposed to 1 where it's a^(b^c). Here, though, 300 is 0 mod 4 and so 2^{300} is 1 mod 5. Then the last digit d is 0 mod 2, and by the Chinese remainder theorem, d is 6.