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Areteem Math Challenge II-A: Number Theory

 
 
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Areteem Math Challenge II-A: Number Theory
by david han - Sunday, 4 June 2017, 11:52 AM
 

Class 6 Q3

why is 17^1999=17^2000*53 ?

 
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Re: Areteem Math Challenge II-A: Number Theory
by David Reynoso - Monday, 5 June 2017, 11:29 AM
 

Hello David!

We have that $53\cdot17 = 901$, this means that $53\cdot 17 \equiv 1 \pmod{100}$ (the remainder of $901$ when dividing by $100$ is $1$), so $53$ is the multiplicative inverse of $17$ modulo $100$. Now, $17^{1999} \equiv 1\cdot 17^{1999} \equiv 53\cdot 17\cdot 17^{1999} \equiv 53\cdot 17^{20000}\pmod{100}$.

Note that this does not mean that $17^{1999}=17^{2000}\cdot53$, but that they have the same last two digits.


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Re: Areteem Math Challenge II-A: Number Theory
by david han - Tuesday, 13 June 2017, 3:27 PM
 

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