An old 1990 problem

Number Theory Level 4

Find the remainder when $2^{1990}$ is divided by 1990.

Source: RMO 1990.

The answer is 1024.

1 solution

Otto Bretscher
Apr 20, 2016

$\lambda(995)=396$ so $2^{1989}=2^{5\times 396+9}\equiv 2^9 \pmod{995}$ so $2^{1990}\equiv 2^{10}=\boxed{1024} \pmod{1990}$

Im an amateur at number theory, could you recommend some texts which would make your solution more understandable to me ?

Silver Vice - 5 years, 1 month ago

I would start with some online resources such as this . These concepts are not deep... you will grasp them quickly.The Carmichael Lambda $\lambda(n)$ of $n$ is simply the smallest positive integer $m$ such that $a^m\equiv 1 \pmod{n}$ whenever $\gcd(a,n)=1$

Otto Bretscher - 5 years, 1 month ago

Is this lambda the same as Euler's totient function?

Puneet Pinku - 5 years ago

@Puneet Pinku – No, it's the Carmichael Lamda Function

Otto Bretscher - 5 years ago

Look up Carmichael's Lambda Function.It's really useful in doing modular arithmetic and speeds up the calculations so that you get such brilliant one-liner solutions as posted above.

Abdur Rehman Zahid - 5 years, 1 month ago

Sir we have to just cancel out the common factor and find the remainder(r) and then again multiply the remainder(r) with the common factor we cancled earlier? is this so ?

Chirayu Bhardwaj - 4 years, 8 months ago

An old 1990 problem

Source: RMO 1990.

The answer is 1024.

1 solution

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