If k is a positive integer, what is the remainder when 13^(4k + 2} + 8

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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Bunuel sir, Can you rectify the question? is it \((13^{4K}+2)+8\) or \((13^4*K+2)+8\)

How the remainder is 7?

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

Posted from my mobile device

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

I've rectified it.

Apologies, I'm getting use to the forum code.

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

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[/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!

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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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.

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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