That's a lot of variables... what do I do?

Let ψ \psi be defined as a function such that for any positive integer n n , ψ ( n ) \psi (n) finds the smallest possible integral value for m 2 m \geq 2 such that n m ( m o d 10 ) = n ( m o d 10 ) n^{m} \pmod{10}=n \pmod{10} . Let P P be the probability that if a random nonnegative integer n n is selected that ψ ( n ) = 5 \psi (n)=5 . If P P can be written as a b \frac{a}{b} where a a and b b are coprime, positive integers, find a + b a+b .


The answer is 7.

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2 solutions

Daniel Liu
Apr 7, 2014

We need only consider when n 0 9 ( m o d 10 ) n\equiv 0\to 9\pmod{10} , so let's check the cases for n = 0 9 n=0\to 9 .

Note that for n = 2 , 3 , 7 , 8 n=2,3,7,8 , we have that n 5 n ( m o d 10 ) n^5\equiv n\pmod{10} :

2 , 4 , 8 , 16 , 3 2 \boxed{2},4,8,16,3\boxed{2}

3 , 9 , 27 , 81 , 24 3 \boxed{3},9,27,81,24\boxed{3}

7 , 49 , 343 , 2401 , 1680 7 \boxed{7},49,343,2401,1680\boxed{7}

8 , 64 , 512 , 4096 , 3276 8 \boxed{8},64,512,4096, 3276\boxed{8}

For all the others, m < 5 m<5 .

Therefore, our probability is 4 10 = 2 5 \dfrac{4}{10}=\dfrac{2}{5} and our desired answer is 2 + 5 = 7 2+5=\boxed{7} .

Perfect! :D

Finn Hulse - 7 years, 2 months ago

It seems to me the question is not well stated. In all cases m=1 or m=0 are the lowest integer values that work.

Since these are trivial, one has to assume the author meant m>1.

Steven Perkins - 7 years, 2 months ago

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Thanks. I have updated the problem accordingly.

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Calvin Lin Staff - 6 years, 5 months ago
Ramiel To-ong
Jun 5, 2015

total percentage of P = 40%

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