Adding in 4D

Probability Level pending

What is the probability that the last digit of a sum of two perfect fourth powers, out of the first 10,000 fourth powers, is 1?

If this probability can be expressed as $\dfrac mn$ , where $m$ and $n$ are coprime positive integers, submit your answer as $m+n$ .

The answer is 29.

1 solution

Marta Reece
Dec 24, 2016

While the last digits of the first 10,000 numbers are evenly distributed

$0, 1, 2, 3, 4, 5, 6, 7, 8, 9$

Last digits of their squares are less so

$0, 1, 4, 9, 6, 5, 6, 9, 4, 1$

And the last digits of the fourth powers (squares of those above) even less

$0, 1, 6, 1, 6, 5, 6, 1, 6, 1$

We are left with only 4 possible digits with lopsided probabilities

0 - 0.1

1 - 0.4

5 - 0.1

6 - 0.4

To get 1 as the last digit of a sum of two of these, we have combinations (with their probabilities again listed)

0, 1 - 0.04

1, 0 - 0.04

5, 6 - 0.04

6, 5 - 0.04

for a total probability of

$0.16=16/100=4/25$

Note that there isn't a "uniform distribution on countably infinite elements". As such, I have edited the problem for clarity.

Calvin Lin Staff - 4 years, 5 months ago

You might want to change the question to ask for m + n from the expression m/n, at the moment it says nothing about that

Loppukilpailija - 4 years, 5 months ago

Thanks. I've made the necessary edits.

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Brilliant Mathematics Staff - 4 years, 5 months ago

Adding in 4D

The answer is 29.

1 solution

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