Question

Find the remainder when (33333 to the power 44444 + 44444 to the power 33333) is divided by 7 . (A.1 B.6 C.2 D.3)

Answer 1

this is it right


\(\large \color{black}{\begin{align}
& \dfrac{33333^{44444} + 44444^{33333}}{7}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

Yes

Answer 3

\(\large \color{black}{\begin{align}
& \dfrac{33333^{44444} + 44444^{33333}}{7}\hspace{.33em}\\~\\
& =\dfrac{6^{44444} + 1^{33333}}{7}\hspace{.33em}\\~\\
& =\dfrac{(-1)^{44444} + 1}{7}\hspace{.33em}\\~\\
& =\dfrac{1 + 1}{7}\hspace{.33em}\\~\\
& =\dfrac{2}{7}\hspace{.33em}\\~\\
& =2\hspace{.33em}\\~\\
\end{align}}\)

Answer 4

btw for which exam is this question

Answer 5

Can you please explain the solution ? This question is for general aptitude exams :)

Answer 6

which part u dont get

Answer 7

first first part how did we get 6 +1?

Answer 8

And also the -1

Answer 9

33333 divide by 7 leaves remainder 6 and
44444 divide by 7 leaves remainder 1

Answer 10

when (a-1) is divide by a then the remainder is -1 for a = natural number greater than 1

Answer 11

Oh got it ! Thank you so much for your help :)