Odd man out

Number Theory Level 5

What is the sum of the first two positive odd integers $n$ such that the $n^\text{th}$ arithmetic derivative of $n$ is non-zero?

For example, the third arithmetic derivative of $n$ , is given by $((n')')'$ .

Clarification: The arithmetic derivative of $n$ is given by:

$n' = 0$ for $n$ = 1 or 0.
$n'$ = 1 if $n$ is prime.
$n' = a'b + b'a$ where $a>1$ and $b>1$ are factors of $n$ .

Note: If $n$ has multiple factors you can choose any pair to get the arithmetic derivative.

(Image courtesy of "Top Lead Generators")

The answer is 42.

1 solution

Geoff Pilling
Apr 9, 2016

The first two odd integers, $n$ , with nonzero $n$ th arithmetic derivatives are 15 (since it has prime factors that add up to a number divisible by 4) and 27 since $27'=27$ so $27''''''''''''''''''''''''''' = 27$ . So the sum is 42.

Odd man out

The answer is 42.

1 solution

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