How I can find out the last two(the rightmost) digit of 3^999?

Question

Is there any general way to solve this types of problem? Please help!

Answer 1

This is a problem in arithmetic "modulo 100", i.e. multiples of 100 can be ignored and only the remainders count.

Prime factors of 100 are 2^2 and 5^2
Euler's phi-function (or Totient function) for 2^2 is 2, and for 5^2 is 20. This is simple, but would take too long to explain here. See the web page below.

The magic function is lambda(100) = least common multiple of phi(2^2) and phi(5^2) = lcm(2, 20) = 20. This is why the remainders go round in a cycle of 20. Some numbers will go round in smaller cycles - 10, 5, or 4. The number 49 will go round in a cycle of 2, and the number 1 just stays where it is.

Anyway, 3^20 = 1 mod 100 is guaranteed, so 3^1000 = 1, and 3^999 = 1 / 3 = 201 / 3 = 67 (adding multiples of 100 to make the division come out right). This is the correct answer no matter whether the cycle length for 3 is actually 20, 10, 5 or 4.

Answer 2

Since 100 and 3 are relatively prime, there is one smallest value of p for which 3^p has last digits 01. It's clear that p is a multiple of 4 and that the smallest is p=20: just look at the binomial formula (80+1)^k for k=5. So 3^1000 finishes by 01. If you take 201 out of 3^1000, you have a multiple of 100 and of 3. Therefore 3^1000 is of the form
300 q +201 and 3^999 is 100 q + 67. That's it.

Answer 3

The only way I know is to write out the last two digits of the powers until i find a pattern.

01, 03, 09, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01

That means it repeats every 20 powers of 3. 999 should then end one of those (20*50-1, note how the first term is 3^0), and therefore end in 67.

Answer 4

Look at the final digit of the first few powers of 3:

3, 9, 27, 81, 243, 729, 2187, 6561, ...

Clearly they go in cycles of four, and 999 = 4*249 + 3

Hence 3^996 is at the end of one of these four-cycles, and ends in 1. 3^999 is the third number in the next four-cycle, and so ends in 7.

Sorry, I didn't notice at first that you said the last two digits.

Continuing the above sequence, we see 3, 9, 27, 81, 243, 729, 2187, 6561,19683, 59049, 177147. If there's a pattern here I can't see it yet, and am very sleepy, so see what you can do with it. I'm going to bed. Maybe.

3, 9, 27, 81, 243, 729, 2187, 6561,19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, ... This time I really do give up!

Answer 5

Let the number be N*100 + ab

where ab are the required digits.

Now 3^5 = 243 = 200 + 43

now 43 * 3 = 129

29*3 = 87 87*3 = 261 61*3 = 183

Continue.

Answer 6

yes because of our base 10 number system, certain powers and multiples of numbers repect their last few digits. do a few of the first powers of three, find a repeating pattern. determine the answer when the exponent is 999

Answer 7

No because I have a 1 and a 0 :(