The question is what is the last digit of 3^99. I found the pattern of 3,5,7,9 for powers of three and bute forced it to get the answer which I don’t have in front of me, but I think is 9 if I recall correctly. Took me about 15 minutes. I am sure there is a 30 second method. Anyone know what it is?
The pattern seems to be 3,9,7,1 then it repeats.
So 3^100 would have 25 of those and the last digit 1.
So 3^99 would be 7.
Is this correct?
I get 7.
I looked at the last digit of different powers of 3:
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
From there it repeats in a cycle of 4. 3^100 would be at the end of a cycle, so it ends in 1. 3^99 is one before that, so it ends in 7.
3[sup]99[/sup] = 171792506910670443678820376588540424234035840667.
So, 7 is the last digit.
That’s what I get for posting a work from memory. Yes, the pattern was 1,3,9,7. Don’t know where the five came from. The same pattern in reverse occurs with 3^x where x is divisible by 9 and that is how I figured it out, by doing that 11 times.
OK, so ryoushi and I have been eliminated. Now for your grand prize question.
Name that number. 171 what.
I got 7. Since you are only concerned with the last digit, once a digit repeats, you have a cycle. In this case the length is four. Just divide 99 by four; ignore the quotent and take the remainder 3. The last digit of 3^99 = the last digit of 3^3.
BTW, this is modular arithmetic base 10. I don’t remember the formula when the base is composite. With a prime modulus p:
n = n^p mod p
Or you could use brute force like Q.E.D did, I suppose.
171 quattuordecillion.
What brute force? I just plugged it into MathCad.
Dr Matrix wrote
If a and m are coprime, then
a[sup][symbol]f/symbol[/sup] = 1 ( mod m)
where [symbol]f[/symbol] is the Euler function.
( G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Theorem 72)
Here m = 10 and [symbol]f/symbol = 4, with the possible a being 1, 3, 7 or 9. So, for each of these a
a[sup]4[/sup] = 1 ( mod 10)
and so, for every k,
a[sup]4k[/sup] = 1 ( mod 10)