Mathematics
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OpenStudy (anonymous):
Determine the last digit of the number (123)^999. Any ideas how?
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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
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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 ?
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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 ^
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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
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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 :)
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OpenStudy (anonymous):
Wow, thanks guys!
ganeshie8 (ganeshie8):
u wlc :)