Find a natural number n no greater than 10 millions so that the first 4 digits of its millionth power (n^1,000,000) are different.

12^1000000 = 1762...

How about doing it without a calculator or paper+pencil?

First four digits are 1, 2, 7, 8 (though not necessarily in that order) Nice problem!

Easy to do on excel but cant do it otherwise - am I missing a trick? and how are the first 4 digits 1,2,7,8 (in whatever order) - 22^ 1m = 4795...

Heres my two bits. 1 million + 1. Use the binomial expansion to show that it is the right answer. I'm sorry i haven't worked out if it really works out but definitely something along those lines should work out. Its a bit late in the evening and i'm feeling too lazy to work it out right now.

I guess if I take 1 million+1, it does the job. Let me prove this. To make life easier, let us denote 1,000,000 by m. Then what I am going to prove is,the first four digits of (1+m)^m are different.We know, for any m, we have,(1+1/m)^m < e < (1+1/m)^(1+m).Thus, for us,e - (1+1/m)^m < (1+1/m)^(1+m) - (1+1/m)^m = (1+1/m)^m times (1/m) < e times (1/m) < 1/ 100000 (Since we have m = 1 million)Now, m^m (e - (1+1/m)^m) < m^m times e - (1+m)^m < m^m times (1/100000)We know e=2.7182...Hence, the first four digists of (1+m)^m will be 2718 because the error is at least 5 digits away.So 1+m (i.e. 1000001) does the job.I hope I didn't make too many errors here. And I am sorry I am not familar with typing formulas with Latex.