Question

How many times does 2 divide 123456789^123456789-1

Answer 1

How many times does 2 divide \[\huge 123456789^{123456789}-1\]

Answer 2

i give up, how many?

Answer 3

give me a sec to think :P

Answer 4

I tried to find a pattern, but I couldn't :P

Give us a formula :)

Answer 5

omg i think i popped a blood vessel in my brain. O.o

Answer 6

my guess is none

Answer 7

ok i take that guess back

Answer 8

it has to atlease one...odd^odd=odd
odd-1=even

Answer 9

guessing wont get you through life :P

Answer 10

worked pretty well so far
(i guess)

Answer 11

lol

Answer 12

the last digit alternates between 1 n 9....after 123456789 times, it will be a 9 again. after minus 1, the last digit will be 8 but dats all i got...

Answer 13

pffftt daz eZ (for some ppl, not me...).

Answer 14

if you try to calculate this, you get an overflow error :]

Answer 15

i mean, on your calculator

Answer 16

and in your brain...

Answer 17

that -1 seems incredibly arbitrary to me, all that does is shift the function down by one unit...but what is a one unit shift in comparison to a super huge exponential rate? Any number, even a number as low as 2, raised to the 123456789th power is going to be reaching exceedingly large values in a short period of time.

Answer 18

-1.000000000 repeating?

Answer 19

Here's the solution!

We could have factored it in this way, but I'll jump straight to just writing this geometric series:

\[\sum_{n=0}^{123456788}123456789^n=\frac{123456789^{123456789}-1}{123456789-1} \]

The sum on the left represents adding together an odd number of odd numbers so the left side is odd. That means the power of 2 that divides \(123456789^{123456789}-1\) is equal to the power of 2 that divides \(123456789-1\).

At this point, you plug this into wolfram alpha and find that it's divisible by 4.

So 2 divides into \(123456789^{123456789}-1\) twice! :P