Question

when 8^263 is expanded out,what digit will be in the units (1's) space?
a)8
b)4
c)2
d)6
e)5 ???

Answer 1

3255866422304616344765552632188114158762089024568314531443485259650408807528140659922574316831813618526821245406949824436469141432675471230646169947427163108444901161872077421124549944292877941762189949285100879873872435565174053364826112

It is a two

Answer 2

thanks

Answer 3

there are a couple of ways to do this. you can find the pattern in the powers.

8^1 gives a ones digit of 8
8^2 gives a ones digit of 4
8^3 gives 2
8^4 gives 6
8^5 gives 8

and it will repeat from there. So we have a sequence like:

8, 4, 2, 6, 8, 4, 2, 6, .... and we want to know what the 263rd term is.
Since this cycles every four terms, we do 263/4 and find its remainder.
\[263 = 4(65)+3\]the remainer is 3, so the 263rd term will land on the third term of the cycle which is 2.

Answer 4

joemath will u help me in a problem?

Answer 5

i can try, whats up?

Answer 6

0/0=undefined
2*0=0
2=0/0 then,2=undefined? but it is impossible

Answer 7

right, thats impossible. because 0/0 is undefined, it wont follow any of our rules we have in math. Its like all those proofs you see for 1 = 0 and the like.

Answer 8

if that were a true statement, then not just 2, but any number would be undefined.
\[a*0 = 0 \Rightarrow a = \frac{0}{0}\]

Answer 9

yes not only 2.so,it does not follow the rules.

Answer 10

right. whenever someone comes to you with some bogus claim like:
\[\sqrt{-4} = 2\]
or whatever (that happened yesterday lol), they are always breaking a rule somewhere, the name of the game is, can you find what they did wrong.

Answer 11

ok.thanks

Answer 12

The only sorta "weird" statement that is a true statement is:
\[.9999\ldots = 1\]
That is true. Other than that though, all bogus! lolol

Answer 13

maths is full of mystery and beauty .....

Answer 14

indeed it is....indeed it is lol.

Answer 15

lol