Question

Find the last three digits of the Number \(2003^{2002^{2001}}\).

Answer 1

This is assuming \(2003^{(2002^{2001})}\)

To find the last three digits, we want to take the number mod 1000. Thus, the question is equivalent to asking what is\[3^{2002^{2001}} \mod 1000\]Also, we know that\(\phi(1000)=400\) So \[3^{400} \equiv 1 \mod 1000\].

Now, we need to find \(2002^{2001} \mod 400\) Using the same method, we have that it's equivalent to \(2^{2001} \mod 400\), and \(\phi(400)=160\). Thus, \(2001 = 12*160 + 81\), so we need to calculate \(2^{81} \mod 400\). Using successive squaring this is relatively easy to do by hand, and you find that \(2^81 \equiv 352 \mod 400\)

Finally, we need to calculate \(3^{352} \mod 1000\). Once again, we can use successive squaring, and we find that \(3^{352} \equiv 241 \mod 1000\).

Answer 2

If we're assuming \((2003^{2002})^{2001}\), the answer is different.

Answer 3

i am going to make a guess that it is the second one

Answer 4

\[2003^{400}\equiv1 (\text{mod} 1000)\] i think

Answer 5

If it's the second one, it's almost trivial. \(2002*2001 = 4006002 = 400k+2\) for some \(k\), so it's \(3^2 \equiv 9 \mod 1000\).

Answer 6

no that is wrong what i wrote

Answer 7

it is 9

Answer 8

Your first statement was correct, since the gcd of 2003 and 1000 is 1, \(2003^{400} \equiv 1 \mod 1000\).

Answer 9

and 9 only leaves remainders of 1 and 9, so answer is 9 if i am thinking clearly

Answer 10

@KingGeorge correct me if i am wrong, but we can use
\[a^{\phi(1000)}\equiv 1 (\text{mod }1000)\] yes?

Answer 11

Correct.

Answer 12

If \(gcd(a, 1000) =1 \)

Answer 13

ah right. guess we need that too, i forgot. but it is true in any case

Answer 14

i mean in this case

Answer 15

It is true in this case.

More formally,\[(2003^{2002})^{2001} \equiv 2003^{4006002}\equiv 3^{10015*400}\cdot 3^2 \mod 1000\]So \[3^{10015*400}\cdot 3^2 \equiv 1*3^2 \equiv 9 \mod 1000\]

Answer 16

short and sweet