Was just talking to a friend who is a Math prof, when she posed me this question: What is the last digit of the number 7 raised to 85?
When you start working that out, 7 raised to 1, ends in 7; raised to 2 ends in 9; raised to 3 ends in 3; and raised to 4 ends in 1; and then the cycle repeats. Therefore for any positive integer n, 7 raised to n ends in 7 when n mod 4 is 1; ends in 9 when n mod 4 is 2; ends in 3 when n mod 4 is 3; and ends in 1 when n mod 4 is 1.
In case of 7 raised to 85, 85 modulo 4, is 1. Hence 7 raised to 85 ends in 7.
Which is fine, but one needs to generalise that for all numbers.
Hence one can ask the question: for a number a^^{n} , where both a and n are natural numbers , what is the probability that a^^{n} ends in 1,2,3,4,5,6,7,8,9, and 0, respectively.
Let us first look at each a separately:
(a) For a = any number ending in 1 (the probability of which in a random sample is 0.1) : a^^{n} will end in 1 all the time (100%)
(b) For a = any number ending in 2 (the probability of which in a random sample is 0.1) : a^^{n} will end in 2,4,8, and 6 , 25% of the time each.
( c ) For a = any number ending in 3 : a^^{n} will end in 3,9,7 and 1, 25% of the time each
(d) For a = any number ending in 4: a^^{n} will end in 4 and 6, 50% of the time each
(e) For a = any number ending in 5: a^^{n} will end in 5 hundred percent of the time
(f) For a = any number ending in 6: a^^{n} will end in 6 hundred percent of the time
(g) For a = any number ending in 7: a^^{n} will end in 7,9,3 and 1, 25% of the time each
(h) For a = any number ending in 8: a^^{n} will end in 8,4,2 and 6, 25% of the time each
(i) For a = any number ending in 9: a^^{n} will end in 9 and 1, 50% of the time each
(j) For a = any number ending in 0 (not being 0, since we defined a and n as Natural Numbers): a^^{n} will end in 0 hundred percent of the time.
Now computing each of the probabilities of a^^{n} ending in :
1: 0.1 x 1 (a) plus 0.1 x 0.25 (c ) plus 0.1 x 0.25 (g) plus 0.1 x 0.5 ( i) = 0.1 x 2 = 20 percent
Applying the same method, we get the probability that a^^{n} ends in
1: is 20 percent
2: is 5 percent
3: is 5 percent
4: is 10 percent
5: is 10 percent
6: is 20 percent
7: is 5 percent
8: is 5 percent
9: is 10 percent
0: is 10 percent
I also ran a simulation in excel with a sample size of 1000 randomly generated a^^{n} . The resulting frequency table closely matches my result, hence I think my answer should be correct.
The frequency table generated through a Monte Carlo:
endnos. | frequency | my answer |
0 | 10.241% | 10% |
1 | 19.177% | 20% |
2 | 4.618% | 5% |
3 | 4.719% | 5% |
4 | 11.044% | 10% |
5 | 10.341% | 10% |
6 | 20.181% | 20% |
7 | 4.217% | 5% |
8 | 5.422% | 5% |
9 | 10.040% | 10% |
Your comments are welcome.
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