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Find the remainder when 24^50 - 15^50 is divided by 13.

 Jul 27, 2019
 #1
avatar+103148 
+2

[24^50 - 15^50 ]  / 13   =  

 

[ 24^25 + 15^25 ] [ 24^25 - 15^25] / 13    = 

 

[ (8 * 3)^25 + (5 * 3)^25 ]  [ (8 * 3)^25 - (5 * 3)^25 )  / 13  =

 

[ 8^25 * 3^25 + 5^25 * 3^25 ]  [ 8^25 * 3^25 - 5^25 * 3^25 ]  / 13   = 

 

[ 3^25  ( 8^25 + 5^25) ]  [ 3^25 (  8^25 - 5^25)] / 13

 

[ 3^50 ] [ 8^25 + 5^25 ] [ 8^25 - 5^25 ]  / 13

 

Note that

(8^5) mod 13  = 8

And by a modular math property, since 5 is a factor of 25....then (8^25)mod 13  = 8

 

Also

(5^5) mod 13  = 5

And......using the same property......(5^25) mod 13  = (5^5) mod 13  =  5

 

So [ 8^25 + 5^25 ] /13   =  [ 8^25 mod 13 + 5^25 mod 13 ] /13  =

 

[ 8mod 13 + 5 mod 13 ]/13  =  [8 + 5] /13  = 13 /13  =  13 mod 13  =  0 = remainder

 

So

[ 3^50] * [ 8^25 + 5^25]  * [ 8^25 - 5^25]

             ____________

                     13

 

Is evenly divisible by 13...so

 

(24*50 - 15^50)   is  evenly divisible by 13.....so.....the remainder   =  0

 

 

cool cool cool

 Jul 27, 2019
edited by CPhill  Jul 28, 2019
 #2
avatar
+1

Since (24^2 - 15^2) mod 13 = 0, which is the same as:
8^2 - 5^2 mod 13 = 0, then:
Every EVEN power of 8 and 5 mod 13 =0
Therefore: 8^50 - 5^50 mod 13 = 0

 Jul 27, 2019
 #3
avatar+23082 
+2

Find the remainder when \(24^{50} - 15^{50}\) is divided by \(13\).

 

\(\begin{array}{|rcll|} \hline &&\mathbf{24^{50} - 15^{50} \pmod{13}} \\\\ && \boxed{24\pmod{13}\equiv 24-13 \\ \equiv 11 \\ \equiv 11-13 \\ \equiv -2\pmod{13} \\ 15\pmod{13}\equiv 15-13 \\ \equiv 2 \pmod{13} } \\\\ &\equiv& (-2)^{50} - 2^{50} \pmod{13} \\ &\equiv & (-1)^{50}2^{50}- 2^{50} \pmod{13} \quad | \quad (-1)^{50}=1 \\ &\equiv & 1\cdot 2^{50}- 2^{50} \pmod{13} \\ &\equiv & 2^{50}- 2^{50} \pmod{13} \\ &\equiv & \mathbf{ 0 \pmod{13} } \\ \hline \end{array}\)

 

laugh

 Jul 28, 2019

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