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14^15 mod 15

 Jun 23, 2017
 #1
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0

14^15 mod 15  = 14. Use the calculator here. First calculate: 14^15 and then press "mod" key followed by 15. You should get 14 as the remainder.

 Jun 23, 2017
 #5
avatar+118608 
+1

I shall try your method:)

14^15 = 155568095557812224155568095557812224mod15 = 155568095557812224*mod15

As you can see this calculator does not work that way.

 

I shall try putting it in properly - and yes this calculator is capable of calculating it straight up

(I had thought the number , i mean 14^15, might be too big but it is not)

mod(14^15,15) = 14       laugh

 

I result is directly from the web2 calc on this posting page :)

Melody  Jun 24, 2017
 #6
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You calculated 14^15 twice and concatenated them together!! On the 2.0Webcalc: 14^15 =

155,568,095,557,812,224 mod 15 = 14.

Guest Jun 24, 2017
 #7
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How do you know she calculated it twice?

It’s obviously pasted it twice, but calculated twice, you can’t know that.

 

Blarney bags always jump to conclusions.

 You know, it would be great if you jumped into a crocodile pit. Really great! 

Guest Jun 24, 2017
 #2
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+1

1415+1=(14+1)(1414-1413+1412-......-14+1)=15*(1414-1413+1412-......-14+1)= 1415+1 is divisible by 15

 

therefore, 1415 mod 15=15-1=14

 

Another way to solve it is this one:

 

\( {14}^{15}={(15-1)}^{15}={15}^{15}*\begin{pmatrix} 15\\ 0 \end{pmatrix} -{15}^{14}*\begin{pmatrix} 15\\ 1 \end{pmatrix} +.....+15*\begin{pmatrix} 15\\ 14 \end{pmatrix}- 1*\begin{pmatrix} 15\\ 15 \end{pmatrix}=15*({15}^{14}*\begin{pmatrix} 15\\ 0 \end{pmatrix}- {15}^{13}*\begin{pmatrix} 15\\ 1 \end{pmatrix}+....+1*\begin{pmatrix} 15\\ 14 \end{pmatrix})-1\)

 Jun 24, 2017
edited by Guest  Jun 24, 2017
edited by Guest  Jun 24, 2017
 #3
avatar+118608 
+1

Here is another way

 

14^1=14=    -1mod15

14^2=196= +1 mod15

14^3=2744 = mod(2744,15) = 14 = -1

114^4 = 38416 = mod(38416,15) = 1

 

We have a pattern

 

\(For\;\;n\in \text{Positive integers}\\ 14^{2n}=+1\\ 14^{2n-1}=-1 \\ so\\ 14^{15}\;\;mod\;15=-1\;\;or\;\;14\)

 Jun 24, 2017
 #8
avatar+26367 
0
 Jun 26, 2017

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