Question

how do i find the units digit of


\(\LARGE \begin{align} \color{black}{
58^{{52}^{98}} \hspace{.33em}\\~\\
}\end{align}\)

Answer 1

it is either 2, 4, 6, or 8 right?

Answer 2

YES

Answer 3

pattern would look like \[8,4,2,6\]

Answer 4

since 52 is divisible by 4, the units place in \(8^{52}\) is \(6\)

Answer 5

\[\overbrace{8}^1\overbrace{4}^2\overbrace{2}^3\overbrace{6}^4\]

Answer 6

repeat the process for \(6^{96}\)

Answer 7

how did \(\LARGE 6^{96}\) came from

Answer 8

i guess a mistake on my part , should be \[6^{98}\]

Answer 9

YES how did that came

Answer 10

because the units digit of \(8^{52}\) is \(6\)

Answer 11

is that much clear, or is that confusing?

Answer 12

\[58^{{52}^{98}} = (10x-2)^{4y} \equiv (-2)^{4y}\equiv 6^y \pmod{10}\]

Answer 13

yes it was clear thnx

Answer 14

that look familier to binomial :D

Answer 15

yes, to simplify matter we can replace 10 by 0 in mod 10

Answer 16

\[58^{{52}^{98}} = (6\times10 - 2)^{{52}^{98}} \equiv (6\times 0-2)^{{52}^{98}} = (-2)^{{52}^{98}} \pmod{10}\]

Answer 17

noticing 52 is divisible by 4 allows you to conclude 52^98 is divisible by 4
so we have 52^98 = 4y :

\[ (-2)^{{52}^{98}} = (-2)^{4y}\]

Answer 18

@misty1212 ,how did i find

\(\huge 2^{{117}^{188}}\)

do i need to find the units digit of \(\huge 2^{117}\) first

Answer 19

@satellite73

Answer 20

either of them should be able to assist you :)

Answer 21

\(\large \begin{align} \color{black}{
\dfrac{2^{117}}{10}
\implies 2\hspace{.33em}\\~\\
and \hspace{.33em}\\~\\
\dfrac{2^{188}}{10} \hspace{.33em}\\~\\
\implies 6\hspace{.33em}\\~\\
}\end{align}\)

but the answer is 2

Answer 22

ok this method should work

\(\LARGE \begin{align} \color{black}{
2^{{117}^{188}} \hspace{.33em}\\~\\
\implies 2^{\frac{117^{188}}{4}} \hspace{.33em}\\~\\
\implies 2^{\frac{1^{188}}{4}} \hspace{.33em}\\~\\
\implies 2^{\frac{1}{4}} \hspace{.33em}\\~\\
\implies 2^{1} \hspace{.33em}\\~\\
\implies 2 \hspace{.33em}\\~\\
}\end{align}\)

Answer 23

we may write \(117\) as \(4k+1\) and reduce

Answer 24

\[\large 2^{{117}^{188}} =2^{(4k+1)^{188}} \]

Answer 25

I am applying strategy of \((mod~~4)\) and them finding the equivalent digit to \(\large 2^{mod}\)

Answer 26

\[\large 2^{{117}^{188}} =2^{(4k+1)^{188}} = 2^{4m+1} = 2*2^{4m} \]

Answer 27

yes that \(6^{n}\) trick is powerful , but various method confuse me

Answer 28

there is a mistake in your method, let me show it :

how do you work the units digit of \(3^{117^{188}}\) using your method ?

Answer 29

first i did \( 117^{188} ~~~~(mod ~~4)\) which was \(
1\)

\(so\\~\\
2^1\)

Answer 30

Actually no, your method works perfectly fine for powers of 2 :)

Answer 31

i misinterpreted earlier :O

Answer 32

it works for 8 and 7

Answer 33

that looks interesting

Answer 34

yes i was just looking to quickly find the \(\huge a^{b^{c}}\)
in case of \(\large (mod ~~10)\)

Answer 35

so i think misty's method was little dubious

Answer 36

i wouldnt doubt misty's method, all methods are giving the same answer... depends on which one you're more happy wid thats all :)

Answer 37

so you have a general strategy for finding the units digit of \(\large a^{b^c}\) for all \(a\) ? xD

Answer 38

yes but i m still confused why didnt it work for

\(\Large 2^{{117}^{188}}\)

Answer 39

yes tactics rather than stratagy

Answer 40

fun debate
http://www.chess.com/forum/view/general/strategy-vs-tactics-is-there-a-difference-indeed

Answer 41

looks like this mod4 strategy (tactic) works for all a ?

Answer 42

can we say
\(\large a^{b^c} \) and \(\large a^{\frac{b^c}{4}}\) will have same units digit for all \(a\) ?

Answer 43

looks like it works !

Answer 44

it should work for a=3 also because 3^4 = 81
since the last digit is 1, it has to work

Answer 45

\(\LARGE \begin{align} \color{black}{
7^{{152}^{188}} \hspace{.33em}\\~\\
\implies 7^{\frac{152^{188}}{4}} \hspace{.33em}\\~\\
\implies 7^{\frac{0^{188}}{4}} \hspace{.33em}\\~\\
\implies 7^{\frac{0}{4}} \hspace{.33em}\\~\\
\implies \dfrac{7^{4}}{10} \hspace{.33em}\\~\\
\implies \frac{1}{10} \hspace{.33em}\\~\\
\implies 1 \hspace{.33em}\\~\\
}\end{align}\)

It will only work for \(2,7,8\) as they have a cycle of \(4\)

Answer 46

for three there is a more easy one !

Answer 47

it works for all \(a\)
i agree we can have more easier methods for 3,4,5,6,9 :)

Answer 48

your method works for all \(a\)

Answer 49

show me a counter example if you can :)

Answer 50

well yes it works for 2,3,7,8 as they have a cycle of 4 but but in case of cycle of 2 there is another lazy tactic!

Answer 51

you want to take mod 2 for the exponent instead of mod 4 ?

Answer 52

\(\large \begin{align} \color{black}{
\normalsize \text{unit's digit of } \hspace{.33em}\\~\\
9^{{152}^{188}} \hspace{.33em}\\~\\
9^{odd}\implies 9 \hspace{.33em}\\~\\
9^{even}\implies 1 \hspace{.33em}\\~\\
{152}^{188} \normalsize \text{its even } \hspace{.33em}\\~\\
9^{{152}^{188}} \implies
9^{even}\implies 1 \hspace{.33em}\\~\\
}\end{align}\)

Answer 53

these will work for \(4~~ and ~~9\)

Answer 54

beautiful xD

for 4 :

4^odd = 4
4^even = 6

is it ?

Answer 55

yes that one!

Answer 56

it saves time haha

Answer 57

Clever!

Answer 58

and no need for \(1,5~~and~~6\)