1. remainder when 2^91 is divided by 7 ?
2. remainder when 25! is divided by 10^7 ?

please explain the approach :) thank you !

Question

1. remainder when 2^91 is divided by 7  ?
2. remainder when 25! is divided by 10^7 ? 

please explain the approach :) thank you !

anonymous · Answer

$2^2=2,2^2=4,2^3$ patter will repeat remainder when divided by 7 of each of these is $2,4,1$

anonymous · Answer

*pattern

anonymous · Answer

ok.. understood so far ! :)

anonymous · Answer

we can check the next ones
$2^4=16$ $2^5=32, 2^6=64$ remainders are again $2,4,1$

anonymous · Answer

so repeats every 3
now $2^{91}$ should be ok right?

anonymous · Answer

answer is 2 ?

anonymous · Answer

i think so yes

anonymous · Answer

thats brilliant :D 
thank you!

anonymous · Answer

since 9 goes in to 90 evenly, the remainder for $2^{90}=1$ and so remainder of $2^{91}$ would be 2

anonymous · Answer

yw

anonymous · Answer

as for the next one i am going to guess at 0 but we need to know how many zeros are in $25!$

anonymous · Answer

no that is wrong, damn

anonymous · Answer

because there are only 6 zeros in $25!$ so you need the last non zero digit
i am not sure how to do this without a direct computation though

anonymous · Answer

the answer is 4 but that is by computing. i can't think of a snap way to do it 
let me know if you come up with one

anonymous · Answer

sure :)