Question

Let k= 2008^2 + 2^2008. What is the units digit of k^2 + 2^k​

Answer 1

Answer:

this is the OE

The units digit of 2^n is 2, 4, 8, and 6 for n = 1, 2, 3, and 4,

respectively. For n > 4, the units digit of 2^n is equal to that of 2n^−4 .

Thus for every positive integer j the units digit of 2^4j is 6,

and hence 2^2008 has a units digit of 6.

The units digit of 2008^2 is 4.

Therefore the units digit of k is 0,

so the units digit of k^2 is also 0.

Because 2008 is even, both 2008^2 and 2^2008 are

multiples of 4.

Therefore k is a multiple of 4, so the units digit of 2^k is 6, and

the units digit of k^2 + 2^k is also 6

Answer 2

Answer:

$\huge\underline\mathfrak\pink{only \: for \: pts}\star$