Find the remainder

Question

Find the remainder when the expression ${( {{{257}^{33}} + 46})^{26}}$ is divided by $50$.

Is there easy method to solve questions of these types?

Answer 1

$$257\equiv7\pmod{50}$$

Now $7^2\equiv-1\pmod{50}$

$7^{33}=7(7^2)^8\equiv7(-1)^8\equiv7\pmod{50}$

$$257^{33}+46\equiv7+46\pmod{50}$$

$$\implies{( {{{257}^{33}} + 46})^{26}}\equiv53^{26}$$

Now $53^{26}=(50+3)^{26}\equiv3^{26}\pmod{50}$

$3^{26}=(3^2)^{13}=(-1+10)^{13}\equiv-1+13\cdot10\pmod{10^2}\equiv29$

Answer 2

$\!\!\bmod{25}\!:\,\ 257^{\large 33}\equiv 7(7^{\large 2})^{\large 16}\equiv 7(-1)^{\large 16}\equiv \color{#c00}7$

so $\ x \equiv (\color{#c00}7\!+\!46)^{\large 26}\!\equiv\! \underbrace{3^{\large 26}\!\equiv (3^{\large 3})^{\large 2}}_{\large \phi(25)\ =\ 20}\!\!\equiv 2^{\large 2}.\ $ Thus $\ x = 2^{\large 2}\! + 25\pmod{\!50}\,$ by $x$ odd

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Remark $ $ We did use the tag Chinese Remainder Theorem = CRT: we computed $\,x\equiv 29\!\pmod{\!50}$ from $\,\underbrace{x\equiv 1\pmod{\!2}}_{\large x\ {\rm is}\, \ \color{#0a0}{\rm odd}}\,$ and $\ x\equiv 4\pmod{\!25}\!\iff\! x\equiv 4+25k,\,$ which is $\color{#0a0}{\rm odd} \iff k$ is odd.