# Remainder?

• Oct 26th 2009, 02:31 AM
a69356
Remainder?
Q) Find the remainder of 37^47^57 divided by 16?

Any help will be appreciated?

Thanks,
Ashish
• Oct 26th 2009, 10:49 AM
Soroban
Hello, Ashsh!

I think I've solved it . . .

Quote:

Find the remainder of $\displaystyle 37^{47^{57}}$ divided by 16
Since $\displaystyle 37 \equiv 5 \text{ (mod 16)}$

. . 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!
.