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# If k is a non negative integer, what is the remainder when 7

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Intern
Joined: 11 Apr 2012
Posts: 36
If k is a non negative integer, what is the remainder when 7  [#permalink]

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Updated on: 20 Nov 2016, 02:34
6
00:00

Difficulty:

25% (medium)

Question Stats:

77% (01:34) correct 23% (02:00) wrong based on 171 sessions

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If k is a non negative integer, what is the remainder when [7^(12k+2)] + 3 is divided by 10 ?

A. 0
B. 1
C. 2
D. 3
E. 4

Originally posted by GMATBaumgartner on 25 Aug 2012, 21:41.
Last edited by HKD1710 on 20 Nov 2016, 02:34, edited 1 time in total.
corrected the format
Director
Joined: 22 Mar 2011
Posts: 598
WE: Science (Education)
Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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25 Aug 2012, 23:30
vinay911 wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

a)0
b)1
c)2
d)3
e)4

In fact, the question asks what is the last digit of the number $$7^{12k+2}+3$$.

The powers of 7 end in 7,9,3,1 cyclically, so, because $$12k+2$$ is a multiple of 4 plus 2, $$7^{12k+2}$$ ends in 9.
In conclusion, $$7^{12k+2}+3$$ ends in 2.

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Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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26 Aug 2012, 10:43
1
1
EvaJager wrote:
vinay911 wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

a)0
b)1
c)2
d)3
e)4

In fact, the question asks what is the last digit of the number $$7^{12k+2}+3$$.

The powers of 7 end in 7,9,3,1 cyclically, so, because $$12k+2$$ is a multiple of 4 plus 2, $$7^{12k+2}$$ ends in 9.
In conclusion, $$7^{12k+2}+3$$ ends in 2.

Better way, it says K is a non negative integer means the answer has to be true for all non negative integers .. weather k is 0 or 987654 ....
Put k = 0 and wrap it up!
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Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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26 Aug 2012, 14:28
mandyrhtdm wrote:
EvaJager wrote:
vinay911 wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

a)0
b)1
c)2
d)3
e)4

In fact, the question asks what is the last digit of the number $$7^{12k+2}+3$$.

The powers of 7 end in 7,9,3,1 cyclically, so, because $$12k+2$$ is a multiple of 4 plus 2, $$7^{12k+2}$$ ends in 9.
In conclusion, $$7^{12k+2}+3$$ ends in 2.

Better way, it says K is a non negative integer means the answer has to be true for all non negative integers .. weather k is 0 or 987654 ....
Put k = 0 and wrap it up!

What if one of the answers, say E, would have been "Cannot be determined" ?
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Intern
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Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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18 Oct 2012, 21:28
GMATBaumgartner wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

A. 0
B. 1
C. 2
D. 3
E. 4

For finding remainders with 5 & 10 : just find the last digit of the eqn

7 has a cyclicity of 4: 7,9,3,1

$$= 7^{12k}*7^{2} + 3$$

7^2 last digit : 9

12*K = Multiple of 4 : hence lats digit : 1

$$(7^{12})^{K}$$ : last digit : 1

= 9*1 + 3

=12

=12/10

R =2
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Joined: 11 Apr 2012
Posts: 36
Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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18 Oct 2012, 22:00
mindmind wrote:
GMATBaumgartner wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

A. 0
B. 1
C. 2
D. 3
E. 4

For finding remainders with 5 & 10 : just find the last digit of the eqn

7 has a cyclicity of 4: 7,9,3,1

$$= 7^{12k}*7^{2} + 3$$

7^2 last digit : 9

12*K = Multiple of 4 : hence lats digit : 1

$$(7^{12})^{K}$$ : last digit : 1

= 9*1 + 3

=12

=12/10

R =2

Hi mindmind,
I am getting units digit as 7 when i try to find $$(7^{12})^{K}$$
not sure where I am going wrong. can you please elaborate on that step ?
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Joined: 23 May 2012
Posts: 28
Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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Updated on: 18 Oct 2012, 22:46
1
1
GMATBaumgartner wrote:
mindmind wrote:
GMATBaumgartner wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

A. 0
B. 1
C. 2
D. 3
E. 4

For finding remainders with 5 & 10 : just find the last digit of the eqn

7 has a cyclicity of 4: 7,9,3,1

$$= 7^{12k}*7^{2} + 3$$

7^2 last digit : 9

12*K = Multiple of 4 : hence lats digit : 1

$$(7^{12})^{K}$$ : last digit : 1

= 9*1 + 3

=12

=12/10

R =2

Hi mindmind,
I am getting units digit as 7 when i try to find $$(7^{12})^{K}$$
not sure where I am going wrong. can you please elaborate on that step ?

$$=7^{12*k} ... 7^{0} .. 7^{12} ....7^{24} ....$$

$$= 7^{4n}$$ .. all of the above are multiples of 4 ....

Cyclicity of 7 :4
1 >> 7
2>>,9
3>>,3,
4>> 1

So , the last digit is 1.

Originally posted by mindmind on 18 Oct 2012, 22:10.
Last edited by mindmind on 18 Oct 2012, 22:46, edited 1 time in total.
Intern
Joined: 11 Apr 2012
Posts: 36
Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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18 Oct 2012, 22:24
ok thanks, I was taking value of k=0,1,2 and getting 0,12,24 ... which on dividing by cyclicity 4 - reminder 0 .Does this also show the same as above(w/o taking it as a multiple of 4) .
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Joined: 06 Sep 2013
Posts: 1630
Concentration: Finance
Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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09 Feb 2014, 07:42
First we have that 7^14 = (10-3)^14, so we stay with (-3)^14/10, now, since exponent is even number will be positive. Then (3^2)^7 / 10 = (10 - 1)^7/10. So we have a remainder of -1 + 3 = 2, when the expression is divided by 10.

We could also have just realized that since its division by 10, the only thing we need is the units digit as the remainder of a number when divided by 10 is the units digit. Therefore 7^14 will have UD of 9 + 3 = 12 UD of 2 / 10 remainder is 2

Hope this clarifies

Cheers
J
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Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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09 Feb 2014, 23:47
GMATBaumgartner wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

A. 0
B. 1
C. 2
D. 3
E. 4

Let us say k = 1 then 7^(14) + 3 will give what remainder. Let us try an find a pattern here in the units digit as it will be the remainder when the number is divided by 10.

7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
So the pattern repeats in the cycle of 4. At 7^12 it will be 1 and 7^14 will be 9.

9 + 3 will give us 12 hence when 12 is divided by 10 the remainder is 2.
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Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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10 Feb 2014, 05:10
Please check the solutions using the units digit posted above. They are more straightforward. Remember than when a number is divided by 10 all we care is the units digit which will be the remainder

Thanks
Cheers
J

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Joined: 24 Oct 2012
Posts: 165
Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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17 Jun 2015, 23:55
GMATBaumgartner wrote:
If k is a non negative integer, what is the remainder when 7^(12k+2)+3 is divided by 10 ?

A. 0
B. 1
C. 2
D. 3
E. 4

My Attempt:

This question basically tests concept of power cycle and Remainder property

7 has power cycle of 4, that means unit digit will follow below patter.
Unit digit of 7^1 -> 7 (4K +1)
Unit digit of 7^2 -> 9 (4K +2)
Unit digit of 7^3 -> 3 (4K +3)
Unit digit of 7^4 -> 1 (4K)

Cycle repeats after 4.
(12k+2)/4 will leave remainder 2 -> (4K + 2 ) patter.
Hence 7 ^ (12k+2) wil have unit digit as 9

Hence 7^(12k+2)+3 will have unit digit as 2.

Hence anything with unit digit as 2 when divided by 10 will leave remainder as 2

Hence Option B is correct
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Re: If k is a non negative integer, what is the remainder when 7  [#permalink]

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