Q) Find the remainder of 37^47^57 divided by 16?
Any help will be appreciated?
Thanks,
Ashish
Hello, Ashsh!
I think I've solved it . . .
Since $\displaystyle 37 \equiv 5 \text{ (mod 16)}$Find the remainder of $\displaystyle 37^{47^{57}}$ divided by 16
. . we have: .$\displaystyle 5^{47^{57}} \text{ (mod 16)}$
Now consider consecutive powers-of-5 (mod 16):
. . $\displaystyle \begin{array}{|c|c|}
2^n & \text{(mod 16)} \\ \hline \\[-4mm]
5^1 & 5 \\ 5^2 & 9 \\ 5^3 & 13 \\ 5^4 & 1 \\ 5^5 & 5 \\ \vdots & \vdots \end{array}$
The remainders step through a 4-step cycle: .$\displaystyle 5-9-13-1- \hdots$
Now we must determine: .$\displaystyle 47^{57}\text{ (mod 4)}$
We have: .$\displaystyle 47^{57}\text{ (mod 4)} \;=\;(\text{-}1)^{57}\text{ (mod 4)} \;=\;-1\text{ (mod 4)}\;=\;3\text{ (mod 4)}$
Hence: .$\displaystyle 5^{47^{57}} \;=\;5^3\text{ (mod 4)} $
And: .$\displaystyle 5^3 \div 16$ has remainder 13.
Therefore: .$\displaystyle 37^{47^{57}} \div 16\,\text{ has remainder }13.$
But someone check my reasoning and work ... please!
.