Victor's Level 5 Challenges - 2

This means $l \equiv m \equiv n = W \pmod {10^4}$ for some positive integer $W$ which we're trying to minimize.

The trick is to find the smallest value of $x$ such that $3^x \equiv 1 \pmod{10^4}$ . By using Euler Totient Function, we have $3^{4000} \equiv 1 \pmod {10^4}$ .

This suggests that $3^{2000} \equiv 1 \text{ or} -1$ , considering modulo $10^4$ . Binomial expansion, only see the last few terms: $3^{2000} = (10 - 1)^{2000} \equiv 1$ .

So $3^{2000} \equiv 1$ only.

Repeat. Binomial expansion: $3^{1000} = (10 - 1)^{500} \equiv { 500 \choose 2 } \cdot 10^2 - 500 \cdot 10 + 1 \equiv 1$

Then, $3^{1000} \equiv 1$ only.

Repeat. Binomial expansion: $3^{500} = (10 - 1)^{500} \equiv { 500 \choose 2 } \cdot 10^2 - 500 \cdot 10 + 1 \equiv 1$

Which yields $3^{500} \equiv 1$

Repeat. Binomial expansion:

$3^{250} = (10-1)^{125} \equiv 1 - { 125 \choose 1} \cdot 10 + { 125 \choose 2 } \cdot 10^2 - {125 \choose 3} \cdot 10^3 \not \equiv 1$

Analogously, We can get $3^{100} \not \equiv 1$

Hence, the smallest value of $x$ is $500$

Thus the last four digits of $3^{500n}$ is equals to $0001$ , for natural number $n$

So the smallest three values are $3^{500}, 3^{1000}, 3^{1500}$ , but $500 + 1000 \not > 1500$ which doesn't satisfy the triangle inequality.

The solution for $(l,m,n)$ is then $(500+Q,1000+Q,1500+Q)$ for positive numbers $Q \geq 1$

Which means the minimum $l+m+n$ equals $(500+1) + (1000+1) + (1500+1) = \boxed{3003}$

I USED THE CONCEPT OF ORDER OF 3 MODULO 625 AND 16 THEY WERE 500, 4 RESPECTIVELY. SO, ORDER OF 3 MODULO 10000 IS 500 . NOW USING TRIANGLE INEQUALITY, I FOUND THAT n IS ATLEAST 501 AND THUS THIS IS SOLVED.

I simply try to see how the last digit/ last two digits... till 4 repeat. Took me a lot. I thought I could do it in c++ but it gave me an error cause it can't get to numbers that high. I used wolfram.