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

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Noshad
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   18 Mar 2019, 03:47
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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
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D
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   18 Mar 2019, 05:34
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chetan2u wrote:
Noshad 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


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
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   18 Mar 2019, 04:53
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Noshad 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 \(10^2+2^{4*something}\) or 0+2^4=0+6=6

Thus D

Editing, took a wrong value of k
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   18 Mar 2019, 05:05
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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
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Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   18 Mar 2019, 05:10
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Noshad 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


Right, I took the value of k wrongly as 10
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Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   27 Nov 2019, 09:00
chetan2u wrote:
Noshad 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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Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   27 Nov 2019, 09:25
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TheNightKing wrote:
chetan2u wrote:
Noshad 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
Thus our answer =0+6=6
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Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   27 Nov 2019, 09:40
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
Thus our answer =0+6=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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Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   28 Nov 2019, 09:05
Noshad 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


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
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Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink] New post   20 Apr 2024, 06:35
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Re: Let k = 2008^2 + 2^2008 . What is the units digit of k^2 + 2^k ? [#permalink]
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