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If k is a positive integer, what is the remainder when 13^(4k + 2} + 8
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dcummins
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If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  Updated on: 15 Sep 2020, 05:44
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If k is a positive integer, what is the remainder when \(13^{4k+2} +8\) is divided by 10 ?

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Originally posted by dcummins on 10 Dec 2018, 19:03.
Last edited by Bunuel on 15 Sep 2020, 05:44, edited 4 times in total.
Edited the question.
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chetan2u
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  11 Dec 2018, 18:39
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hibobotamuss wrote:
I don't quite get this. Can someone explain a little better? Bunuel chetan2u



Hi...
When you divide any number by 10, the remainder is the units digit..
WHY?..
Because the closest multiple of 10 to that number will be the last number having units digit as 0. So whatever extra is there in units digit is the remainder.

So we have to find the units digit of the term.
All digits have a cylicity when it comes to units digit...
3 gives 3,9,7,1,3,9,7,1,3,9.....
\(3^1=3;3^2=9;3^3=27,3^4=81;3^5=243\)..unit digits are 3,9,7,1,3...
Also any number having 3 as units digit will always have same cylicity as 3, so let it be 1876543 or just 3, both will follow same cylicity...
Therefore \(13^{4k+2}\) will have same units digit as 13^2, so units digit will be 9..
So the units digit of entire term will be 9+8=17, thus 7 will be the remainder.

A
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  10 Dec 2018, 19:17
Bunuel sir, Can you rectify the question? is it \((13^{4K}+2)+8\) or \((13^4*K+2)+8\)
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  10 Dec 2018, 19:47
How the remainder is 7?
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  10 Dec 2018, 19:50
sghoshgt wrote:
How the remainder is 7?


sghoshgt, exactly, shouldn't the remainder be 1?

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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  Updated on: 12 Jun 2019, 13:24
To solve this problem examine the functioning of the units digit of \(13^4^k+2\)

Any integer >10 we can refer to the units digit to identify the pattern for subsequent powers of that digit as follows:
\(13^1= 13\)

\(13^2 = 169\)

\(13^3 = 2197\)

\(13^4 = 28,561\)

This is the 3s units digit pattern
\(3-9-7-1\)

It has a cyclicity of 4, meaning its unit digit repeats each multiple of 4.

Thus units digit of 3-9-7-1-3-9 add 8 = 17
17 --> Units digit of 7
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Originally posted by dcummins on 10 Dec 2018, 19:55.
Last edited by dcummins on 12 Jun 2019, 13:24, edited 3 times in total.
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  10 Dec 2018, 19:57
Afc0892 wrote:
Bunuel sir, Can you rectify the question? is it \((13^{4K}+2)+8\) or \((13^4*K+2)+8\)


I've rectified it.

Apologies, I'm getting use to the forum code.
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  10 Dec 2018, 20:09
dcummins wrote:
To solve this problem examine the functioning of the units digit of \(13^4^k+2\)

Any integer >10 we can refer to the units digit to identify the pattern for subsequent powers of that digit as follows:
\(13^1= 13\)

\(13^2 = 169\)

\(13^3 = 2197\)

\(13^4 = 28,561\)

This is the 3s units digit pattern
\(3-9-7-1\)

It has a cyclicity of 4, meaning it repeats each multiple of 4.

Thus units digit of 3-9-7-1-3-9 add 8 = 17
1[/m]7 --> Units digit of 7


The question is 13 Whole to the power 4k+2 and not just \(13^{4k} + 2\)?
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  10 Dec 2018, 20:59
Afc0892 wrote:
dcummins wrote:
To solve this problem examine the functioning of the units digit of \(13^4^k+2\)

Any integer >10 we can refer to the units digit to identify the pattern for subsequent powers of that digit as follows:
\(13^1= 13\)

\(13^2 = 169\)

\(13^3 = 2197\)

\(13^4 = 28,561\)

This is the 3s units digit pattern
\(3-9-7-1\)

It has a cyclicity of 4, meaning it repeats each multiple of 4.

Thus units digit of 3-9-7-1-3-9 add 8 = 17
1[/m]7 --> Units digit of 7


The question is 13 Whole to the power 4k+2 and not just \(13^{4k} + 2\)?



I clarified this in the stem. Re-read it please
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  11 Dec 2018, 07:09
I don't quite get this. Can someone explain a little better? Bunuel chetan2u
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  26 Dec 2018, 14:29
Hi...
When you divide any number by 10, the remainder is the units digit..
WHY?..
Because the closest multiple of 10 to that number will be the last number having units digit as 0. So whatever extra is there in units digit is the remainder.

So we have to find the units digit of the term.
All digits have a cylicity when it comes to units digit...
3 gives 3,9,7,1,3,9,7,1,3,9.....
\(3^1=3;3^2=9;3^3=27,3^4=81;3^5=243\)..unit digits are 3,9,7,1,3...
Also any number having 3 as units digit will always have same cylicity as 3, so let it be 1876543 or just 3, both will follow same cylicity...
Therefore \(13^{4k+2}\) will have same units digit as 13^2, so units digit will be 9..
So the units digit of entire term will be 9+8=17, thus 7 will be the remainder.

A[/quote]

Thank you for your answer!

I just have one question, how do we know that the cyclicity number that we have to choose is 9?

Kind regards!
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  26 Dec 2018, 18:18
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jfranciscocuencag wrote:
Hi...
When you divide any number by 10, the remainder is the units digit..
WHY?..
Because the closest multiple of 10 to that number will be the last number having units digit as 0. So whatever extra is there in units digit is the remainder.

So we have to find the units digit of the term.
All digits have a cylicity when it comes to units digit...
3 gives 3,9,7,1,3,9,7,1,3,9.....
\(3^1=3;3^2=9;3^3=27,3^4=81;3^5=243\)..unit digits are 3,9,7,1,3...
Also any number having 3 as units digit will always have same cylicity as 3, so let it be 1876543 or just 3, both will follow same cylicity...
Therefore \(13^{4k+2}\) will have same units digit as 13^2, so units digit will be 9..
So the units digit of entire term will be 9+8=17, thus 7 will be the remainder.

A


Thank you for your answer!

I just have one question, how do we know that the cyclicity number that we have to choose is 9?

Kind regards![/quote]

Hi ..
We know cylicity is 3,9,7,1,3,9,...
So the cylicity is in 4s...(3,9,7,1),(3,9,7,1)....
So the cylicity is same for first, fifth, ninth terms.... And for second, sixth, tenth and so on.
That is every fourth term is same...
Here we have 13^{4k+2}...
Now 4k means every fourth term, that is 13^{4k} will have _,_,_,1,_,_,_,1...that is it will have 1..
But we are looking for (4k+2)th term and it will have same number as the 2nd term, and, therefore, it is 9...
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  26 Dec 2018, 18:26
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chetan2u wrote:
jfranciscocuencag wrote:
Hi...
When you divide any number by 10, the remainder is the units digit..
WHY?..
Because the closest multiple of 10 to that number will be the last number having units digit as 0. So whatever extra is there in units digit is the remainder.

So we have to find the units digit of the term.
All digits have a cylicity when it comes to units digit...
3 gives 3,9,7,1,3,9,7,1,3,9.....
\(3^1=3;3^2=9;3^3=27,3^4=81;3^5=243\)..unit digits are 3,9,7,1,3...
Also any number having 3 as units digit will always have same cylicity as 3, so let it be 1876543 or just 3, both will follow same cylicity...
Therefore \(13^{4k+2}\) will have same units digit as 13^2, so units digit will be 9..
So the units digit of entire term will be 9+8=17, thus 7 will be the remainder.

A


Thank you for your answer!

I just have one question, how do we know that the cyclicity number that we have to choose is 9?

Kind regards!


Hi ..
We know cylicity is 3,9,7,1,3,9,...
So the cylicity is in 4s...(3,9,7,1),(3,9,7,1)....
So the cylicity is same for first, fifth, ninth terms.... And for second, sixth, tenth and so on.
That is every fourth term is same...
Here we have 13^{4k+2}...
Now 4k means every fourth term, that is 13^{4k} will have _,_,_,1,_,_,_,1...that is it will have 1..
But we are looking for (4k+2)th term and it will have same number as the 2nd term, and, therefore, it is 9...[/quote]

+Kudos

Thank you very much!!! So clear now.
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Re: If k is a positive integer, what is the remainder when 13^(4k + 2} + 8 [#permalink] New post  12 Jun 2019, 09:38
13^(4k+2)=13^4k.13^2
Since it is divided by 10 so we need to consider just the units digit.
13^4k (3,9,7,1-cyclicity) will always have the units digit as 1 and 13^2 has a units digit as 9.
So ((1*9)+8)/10=7

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