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

Question

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

(A) 0

(B) 2

(C) 4

(D) 6

(E) 8

chetan2u · Answer

Easiest way would be to find the units digit of k...
 k = 2008^2 + 2^{2008} , the units digit will be same as that of  8^2+2^{4*502} , that is 4+6 = 10
thus the units digit of  k^2 + 2^k  will be  10^2+2^{4*something}  or 0+2^4=0+6=6

Thus D

Editing, took a wrong value of k

Noshad · Answer

Hi  chetan2u 
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

chetan2u · Answer

Right, I took the value of k wrongly as 10

Manas1212 · Answer

Hi  chetan2u ,  Noshad

We can easily conclude that the unit digit of K = 2008^2 + 2^2008 = 0

but we have to conclude the remainder K leaves when divided by 4 so as to find out the cyclicity(we can not simply conclude it as 10 as 20 having 0 as it's unit digit leaves remainder 0 as compared to 10 leaving 2)

So unit digit of K is 0 , the unit digit of  k^2 = 0---------(a)

2^(2008^2 + 2^2008)
=2^(2008^2) * 2^(2^2008) 
=2^(value divisible by 4) * 2^(value divisible by 4)
=4*4=16
unit digit 6----- (b)

So the unit digit of  k^2 + 2^k  = 0+6 =6

TheNightKing · Answer

chetan2u 
Can you help me understand your highlighted portion. I figure unit digit of k is 0. But I could not figure out what is actually asked

chetan2u · Answer

Hi
That is the post of other member, where he had tagged me..

Anyways..
You have got the point that the units digit of k is 0, so k^2 will also give units digit as 0.

Thus units digit of  k^2+2^k  will depend only on units digit of 2^k..
Now   k = 2008^2 + 2^2008=2^21004^2+(2^2)*2^2006=4(1004^2+2^2006) , which means k is a multiple of 4..let it be k=4x
 2^k=2^{4x}  When we raise 2 to power that is a multiple of 4, the units digit will be same as that of 264=16 or 6
Thus our answer =0+6=6

TheNightKing · Answer

Yes I knew. But since his explanation was more inline with my doubt so quoted that one.

Yup. Now it makes sense. I didn't pay attention to the cyclicity that is required in the second half of the equation. Thank you!

Kinshook · Answer

Unit digit of k = 4 + 6 = 0
Unit digit of k^2 = 0

k is a multiple of 4; k = 4m 
Unit digit of 2^k = 6

Unit digit of k^2 + 2^k = 0 + 6 = 6

IMO D

BrushMyQuant · Answer

k = 2008^2 + 2^{2008}

=> Units' digit of k = Units' digit of  2008^2  + Units' digit of  2^{2008}  = Units' digit of  8^2  + Units' digit of  2^ {2008}   = 4 + Units' digit of  2^ {2008}

Units' digit of  2^ {2008}

Lets start by finding the cyclicity of units' digit in powers of 2

2^1  units’ digit is 2
 2^2  units’ digit is 4 
 2^3  units’ digit is 8 
 2^4  units’ digit is 6
 2^5  units’ digit is 2

That means that units digit of power of 2 has a cycle of 4

=> We need to divide the power (2008) by 4 and check what is the remainder
2008 divided by 4 gives 0 remainder

=> Units digit of  2^{2008}   = Units digit of  2^4  = 6

=> Units' digit of k = Units' digit of (4 + 6) = 0

Units digit of  k^2 + 2^k  = Units digit of (k that ends with 0)^2 + Units digit of 2^(a Multiple of 4)
 k = 2008^2 + 2^ {2008}  = Multiple of 4 + 2^Even power = Multiple of 4]

=> Units digit of  k^2 + 2^k   = 0 + Units digit of  2^{Cycle}  = 0 + Units digit of 2^4 = 0 + 6 = 6

So,  Answer will be D 
Hope it helps!

Link to Theory for Last Two digits of exponents here    .

Link to Theory for Units' digit of exponents here    .

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

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

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