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Mathematics 16 Online
OpenStudy (anonymous):

Determine the last digit of the number (123)^999. Any ideas how?

ganeshie8 (ganeshie8):

log ?

OpenStudy (anonymous):

can u use a calculator ?

OpenStudy (anonymous):

Oh! There's actually no log, I've just changed it to 123^999

OpenStudy (anonymous):

Sorry

ganeshie8 (ganeshie8):

lol okay

OpenStudy (anonymous):

haha , so using mod ?

OpenStudy (anonymous):

Calculator can't solve raise to 999s. lol

ganeshie8 (ganeshie8):

il give a hint : last digit is the remainder u get when u divide the number by 10

OpenStudy (anonymous):

@jricafort yeah ik , i thought ur qs is log .... which can be solved using cal

OpenStudy (anonymous):

so why dont u try to get a pattern? like 123^1 .... 123^n does that work ?

OpenStudy (anonymous):

@ganeshie8 Another hint perhaps?

ganeshie8 (ganeshie8):

second method : look for a pattern as Bswan is guiding u

OpenStudy (anonymous):

123^1 give 3 123^2 give 9 123^3 give 7 123^4 give 1 123^5 give 3 123^6 give 9 and so on :3

ganeshie8 (ganeshie8):

\[123^{999} \equiv 3^{999} \equiv 3 (-1)^{449} \equiv -3 \equiv 7 \mod 10\]

ganeshie8 (ganeshie8):

7 is correct Bswan ^

OpenStudy (anonymous):

u have pat each 4 take 999 mod 4=3 so take case 3 i think 7 :)

OpenStudy (anonymous):

Nice! Got it.. @ganeshie8 & @Bswan Thank you! very helpful! :-)

OpenStudy (anonymous):

:D i was guessing :P

ganeshie8 (ganeshie8):

u could try binomial thm also for doing these kindof probs..

OpenStudy (anonymous):

@ganeshie8 isnt that would be too much for kids ? :O

ganeshie8 (ganeshie8):

By binomial thm, \(123^{999} = 10K + 3^{999}\)

ganeshie8 (ganeshie8):

Bswan that was only to confirm ur answer, not for the OP :)

ganeshie8 (ganeshie8):

applying binomial thm again on 3^999 gives u : \(3^{999} = 10M + 7\)

ganeshie8 (ganeshie8):

you got 3 methods already... use whichever is easy for u :)

OpenStudy (anonymous):

cool :)

OpenStudy (anonymous):

Wow, thanks guys!

ganeshie8 (ganeshie8):

u wlc :)

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