(a) Find the remainders when 250 and 4165 are divided by 7.(b) What is the remainder when...

Also recall that \(2^{2} \equiv 4(\bmod 7)\).

$$ \text { So, } 2^{48} \cdot 2^{2} \equiv 1 \cdot 4(\bmod 7) \begin{aligned} \text { since, } a & \equiv b(\bmod n), c \equiv d(\bmod n) \\ & \Rightarrow a c \equiv b d(\bmod n) \end{aligned} $$

Implies that \(2^{50} \equiv 4(\bmod 7)\) since \(a^{m} \cdot a^{n}=a^{m+n}\)

Hence, when \(2^{50}\) is divided by 7 , the remainder is 4 .

To find the remainder when \(41^{65}\) is divided by 7 as shown below:

Recall that \(41 \equiv 6(\bmod 7)\)

So, \(41^{65} \equiv 6^{65}(\bmod 7)\) since \(a \equiv b(\bmod n) \Rightarrow a^{k} \equiv b^{k}(\bmod n)\)

Implies that \(41^{65} \equiv(-1)^{65}(\bmod 7)(\bmod 7)\) since \(6 \equiv-1(\bmod 7)\)

Thus, \(41^{65} \equiv-1(\bmod 7)(\bmod 7)\) since \((-1)^{65}=-1\)

Therefore, \(41^{65} \equiv 6(\bmod 7)\) since \(6 \equiv-1(\bmod 7)\)

Hence, when \(41^{65}\) is divided by 7, the remainder is 6 .

(b)

The objective is to find the remainder when the sum \(1^{5}+2^{5}+3^{5}+\ldots+99^{5}+100^{5}\) is divided by 4 .

Recall that for any \(x(50+x)^{5}+(50-x)^{5}=2 \cdot 50^{5}+20 \cdot 50^{3} \cdot x^{2}+2 \cdot 5 \cdot 50 \cdot x^{4} \ldots \ldots\) (1)

And if an integer is having two zeroes at the right most, then the integer is divisible by \(4 .\)

So, \(2 \cdot 50^{5} \equiv 0(\bmod 4) \ldots \ldots(2)\)

Since 4 divides 20 ,

It follows that \(20 \cdot 50^{3} \cdot x^{2} \equiv 0(\bmod 4) \ldots \ldots\) (3)

Since 4 divides 500 ,

It follows that \(2 \cdot 5 \cdot 50 \equiv 0(\bmod 4) \ldots \ldots(4)\)

Putting \(1 \leq x \leq 50\) in (1) and using (2), (3), and (4), obtain

\(1^{5}+2^{5}+3^{5}+\ldots+49^{5}+51^{5}+\ldots+100^{5} \equiv 0(\bmod 4)\)

Therefore, \(1^{5}+2^{5}+3^{5}+\ldots+49^{5}+50^{5}+51^{5}+\ldots+100^{5} \equiv 50^{5}(\bmod 4)\) see that \(50^{5}\) is added on

either side of the above congruence.

Further, \(50 \equiv 2(\bmod 4)\)

$$ 50^{5} \equiv 2^{5}(\bmod 4) \text { Since } a \equiv b(\bmod n) \Rightarrow a^{k} \equiv b^{k}(\bmod n) $$

\(50^{5} \equiv 32(\bmod 4)\)

$$ 50^{5} \equiv 0(\bmod 4) \text { Since } 32 \equiv 0(\bmod 4) $$

Using this in the above congruence, obtain

$$ 1^{5}+2^{5}+3^{5}+\ldots+49^{5}+50^{5}+51^{5}+\ldots+100^{5} \equiv 0(\bmod 4) $$

Hence, when the sum \(1^{5}+2^{5}+3^{5}+\ldots+99^{5}+100^{5}\) is divided by 4 , the remainder 0 .