Question

what is the units digits of 943^29

Answer 1

943 x itself by 29 times is what it is asking.

Answer 2

Do you wish for it in expanded form?...

Answer 3

Does this make sense?
943^2 last digit 9
943^3 7
943^4 1
943^5 3

So 943^9 last digit 3

can we continue this to power 29?

Answer 4

\(\large \color{black}{\begin{align}
& \dfrac{943^{29}}{10} \hspace{.33em}\\~\\
& \implies \dfrac{3^{29}}{10} \hspace{.33em}\\~\\
& \implies \dfrac{3\cdot 3^{28}}{10} \hspace{.33em}\\~\\
& \implies \dfrac{3\cdot 9^{14}}{10} \hspace{.33em}\\~\\
& \implies \dfrac{3\cdot (-1)^{14}}{10} \hspace{.33em}\\~\\
& \implies \dfrac{3\cdot 1}{10} \hspace{.33em}\\~\\
& \implies 3 \hspace{.33em}\\~\\
\end{align}}\)

Answer 5

yes
My way would aaalso give 3

Answer 6

@mathmath333

I understand @welshfella 's method, but I'm trying to understand your method. Why did you replace 9 with -1? Also, why did you change 3^(28) to 9^(14)?

Answer 7

the last part ;
3^28 = (3^2)^14 = 9^14

Answer 8

In terms of modular arithmetic when 9 is divided by 10 the remainder is either 9 or -1.

Answer 9

Welshfella, I already knew that. I was just wondering if there was a specific reason to change the base.

Mathmath33, that makes sense! Thanks to both of you!