Q) Find the remainder obtained when 43^101 + 23^101 is divided by 66?
Thanks,
Consider the remainder when divided by $\displaystyle 11$ and the remainder when divided by $\displaystyle 6$.
That is look at:
$\displaystyle 43^{101}=(11\times 4-1)^{101} = k\times 11-1$
for some $\displaystyle k\in \mathbb{N}$
Now do something similar for $\displaystyle 23$ ..
(MOdular arithmetic to base $\displaystyle 11$ and $\displaystyle 6$ will also work if you are familiar with it)
CB
Altought we can use the relation
$\displaystyle
N^{s+1}\equiv N \quad mod \quad Denominator [B_s]
$
Bs is the s-th Bernoulli Number
(100-th Bernoulli Denominator is $\displaystyle 66 \cdot 505$)
(Look Bernoulli Numbers Fundamental Theorem of Arithmetic (in google) for details),
A simpler approach is:
$\displaystyle
43^{101}+23^{101}=(43+23)(43^{100}-43^{99}23+....-43\cdot23^{99}+23^{100})
$
Look here for this factorization
The last expression obviously is divisible by 66, so:
$\displaystyle
43^{101}+23^{101}\equiv 0 \quad mod\quad 66
$