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Senior Manager  P
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Joined: 02 Nov 2018
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 33% (02:08) correct 67% (02:11) wrong based on 64 sessions

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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
Manager  G
Joined: 31 Jul 2017
Posts: 86
Location: India
Schools: Anderson '21, LBS '21
Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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chetan2u wrote:
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

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 $$10^2+2^{10}$$ or 0+2^2=0+4=4

Thus C

OA is wrong

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
Math Expert V
Joined: 02 Aug 2009
Posts: 8336
Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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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

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
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Senior Manager  P
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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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
Math Expert V
Joined: 02 Aug 2009
Posts: 8336
Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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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

Right, I took the value of k wrongly as 10
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GMAT 1: 600 Q46 V27 Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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chetan2u wrote:
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
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
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Math Expert V
Joined: 02 Aug 2009
Posts: 8336
Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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TheNightKing wrote:
chetan2u wrote:
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
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

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
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GMAT 1: 600 Q46 V27 Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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

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

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!
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GMAT 1: 690 Q50 V34 Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?  [#permalink]

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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

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 Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ?   [#permalink] 28 Nov 2019, 09:05
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