Question

whats the units digit of (1273)^122!

Answer 1

so basically you're trying to find the remaidner when divided by 10

Answer 2

yes

Answer 3

units digit = \(\large (1273)^{122!} \mod 10\)

Answer 4

you know that \(1273 \equiv 3 \mod 10\), righgt ?

Answer 5

yes

Answer 6

units digit = \(\large (1273)^{122!} \mod 10\)

\(\large \equiv (3)^{122!} \mod 10\)

yes ?

Answer 7

yes

Answer 8

any ideas what to do next ?

Answer 9

reducing 122!

Answer 10

exactly !
can we write 122! as 2*(122!/2) ?

Answer 11

yes

Answer 12

units digit = \(\large (1273)^{122!} \mod 10\)

\(\large \equiv (3)^{122!} \mod 10\)

\(\large \equiv (3)^{2*122!/2} \mod 10\)

\(\large \equiv (3^2)^{122!/2} \mod 10\)

\(\large \equiv (9)^{122!/2} \mod 10\)

Answer 13

still yes ? :)

Answer 14

oh yes

Answer 15

and you know that \(9 \equiv -1 \mod 10\)

Answer 16

units digit = \(\large (1273)^{122!} \mod 10\)

\(\large \equiv (3)^{122!} \mod 10\)

\(\large \equiv (3)^{2*122!/2} \mod 10\)

\(\large \equiv (3^2)^{122!/2} \mod 10\)

\(\large \equiv (9)^{122!/2} \mod 10\)

\(\large \equiv (-1)^{122!/2} \mod 10\)

Answer 17

+1

Answer 18

Yes! why +1 and not -1 ?

Answer 19

i mean final ans as (-1)power even is +1

Answer 20

Perfect !

Answer 21

units digit = \(\large (1273)^{122!} \mod 10\)

\(\large \equiv (3)^{122!} \mod 10\)

\(\large \equiv (3)^{2*122!/2} \mod 10\)

\(\large \equiv (3^2)^{122!/2} \mod 10\)

\(\large \equiv (9)^{122!/2} \mod 10\)

\(\large \equiv (-1)^{122!/2} \mod 10\)

\(\large \equiv 1 \mod 10\)

Answer 22

The remainder of given number when divided by 10 is 1, so the units digits would be 1

Answer 23

if you want both units digit and hundred's digit, you simply need to work mod 100

Answer 24

why didnt u take (3^3)^(122!/3)

Answer 25

you can take, there is absolutely no problem... lets take it and work :)

Answer 26

\(\large \equiv (3)^{3*122!/3} \mod 10\)

\(\large \equiv (3^3)^{122!/3} \mod 10\)

\(\large \equiv (27)^{122!/3} \mod 10\)

\(\large \equiv (-3)^{122!/3} \mod 10\)

\(\large \equiv (3)^{122!/3} \mod 10\)

we got the SAME thing we have started with, so we just completed one round of running in circle ;)

Answer 27

3^2 gave us "-1" in mod 10
so taking 3^2 is more useful than taking 3^3, getting ?

Answer 28

however it doesn't matter what u take or how u do it for the final answer

Answer 29

ok

Answer 30

we could have taken 3^4 or 3^122 or anything we wish
but 3^2 is the most logical and simplest path to reduction

Answer 31

its tricky

Answer 32

in a way yes, but keep this in mind :
we always look for +1 or -1 in reduction

Answer 33

because (+1)^anything gives you +1 itself

Answer 34

u mean we want 9 or 11

Answer 35

yes exactly !

9 is same as -1
11 is same as +1

in mod 10 ofcourse

Answer 36

okk

Answer 37

got it

Answer 38

we want them because they reduce the huge powers to +1 or -1

Answer 39

ya thns