1. ## Remainder problem

Q) Find the remainder obtained when 43^101 + 23^101 is divided by 66?

Thanks,

2. Originally Posted by a69356
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

3. 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:

$\displaystyle 43^{101}+23^{101}\equiv 0 \quad mod\quad 66$

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### 43^101 23^101÷33

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