9. What digit appears in the units place in the number obtained when 2^320 is multiplied out? A. 0 B. 2 C. 4 D. 6 E. 8
If we wanted the last digit of 2 999 , note that 996 is a multiple of 4 . So at 996 we get a 6 . Now count forward: 2 , 4 , 8 : the answer is 8 . Or else 1000 is a multiple of 4 . Go backwards one step from 6 : the last digit is 8 .
That's a quote from : http://math.stackexchange.com/questions/177312/what-digit-appears-in-unit-place-when-2320-is-multiplied-out
Do you understand?
well im still loading the concept ....
\[2^1=2\]\[2^2=4\]\[2^3=8\]\[2^4={16}\]\[2^5=32\]\[2^6=64\]\[2^7=128\]\[2^8=256\]\[\dots\] we notice last digit cycles, 2,4,8,6,2,4,8,6,2,4,6... the pattern repeats every 4 terms
oh ok i c that ...
right that's helpful
so the last digit will be the same for \[2^{320},2^{316},...\]
uh huh ...
\(320\) is 4 times 79 + 4 \(2^{320}\) has the same last digit as \(2^4\)
right ....thanx alot
Join our real-time social learning platform and learn together with your friends!