Show that the last four digits of $2013^k$ are 0001

Question

Show that a natural $k \ge 1$ exists s.t the last four digits of $2013^k$ (written as a decimal) are 0001.

I understand that k must be of the form k=4m. The last digit of 2013 is 3 and only when powered by multiply of 4 the result ends with 1.

Unfortunately, I don't know how to advance...

Please help, thanks!

The problem, as stated, requires showing only that such such a value of $k$ exists. Are you required to further explicitly give a specific value of $,k,$ that works? — Bill Dubuque, Mar 20 '14 at 17:51

Yiyuan Lee · Accepted Answer · 2014-03-20T23:48:49.353

13

If $2013^k$ has last 4 digits as $1$, then $$2013^k \equiv 1 \pmod {10000}$$

Since $2013$ and $10000$ are coprime, Euler's Theorem asserts that the above relation will hold true for $k = \phi(10000) = 10000\cdot\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{5}\right) = 4000$. In other words,

$$2013^{4000} \equiv 1 \pmod{10000}$$

In fact, Carmichael's Theorem provides a tighter result. It asserts that $k = \lambda(10000) = 500$ will work as well, i.e.

$$2013^{500} \equiv 1 \pmod{10000}$$

edited Mar 20 '14 at 23:48

answered Mar 20 '14 at 14:25

Yiyuan Lee

14,435

Where $\phi(10000)=(1/2)(4/5)(10000)=4000$. – mjqxxxx Mar 20 '14 at 14:33
An since it is a factor of $4000$, you can always use $4000$. That may not be minimal, but you were only asked to prove existence. – Ross Millikan Mar 20 '14 at 14:58

Show that the last four digits of $2013^k$ are 0001

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