If k = 2n - 1, where n is an integer, what is the remainder of k^2/8?
E. Cannot be determined from the information given.
This one can be done very easily with number picking. As you correctly noted k = 2n - 1
means that k is an odd number (basically k = 2n - 1
is a formula of an odd number).
Now let's try several odd numbers:
k=1 --> k^2=1 ---> remainder upon division of 1 by 8 is 1;
k=3 --> k^2=3 ---> remainder upon division of 9 by 8 is 1;
k=5 --> k^2=25 ---> remainder upon division of 25 by 8 is 1;
At this point we can safely assume that this will continue for all odd numbers.
But if you want algebraic approach, here you go:k = 2n - 1
--> what is a remainder when 4n(n-1)+1
is divided by 8:
Now either n
will be even so in any case 4n(n-1)=4*odd*even=multiple \ of \ 8
, so 4n(n-1)
is divisible by 8, so 4n(n-1)+1
divided by 8 gives remainder of 1.
Hope it's clear.