Logarithmic Derivatives of Numbers

Number Theory Level 5

The number-theoretical derivative $Dn$ of a natural number $n$ is defined recursively by the rule $Dp = 1\ \text{for prime}\ p;\ \ \ D(a\cdot b) = a\cdot Db + Da\cdot b.$ (Note the analogy to the product rule for derivatives in calculus.)

From this definition it follows, for instance, that if $n = p^a$ , then $Dn = ap^{a-1}$ . An other concrete example is $D(18) = D(2\cdot 9) = 2\cdot D9 + D2\cdot 9 \\ = 2\cdot 6 + 1\cdot 9 = 12 + 9 = 21.$

We take number-theoretical derivatives to the next level and define the logarithmic derivative $Ln = \frac{Dn}n.$ (We call this "logarithmic derivative" based on the Calculus equality, $d(\ln f(x))/dx = (df(x)/dx)/x$ .)

For instance, $L(18) = \frac{D(18)}{18} = \frac{21}{18} = \frac{7}{6}.$

Question: What are the two smallest distinct natural numbers $a, b$ for which $La = Lb$ ? Give your answer as the sum $a + b$ .

The answer is 31.

1 solution

Arjen Vreugdenhil
Feb 2, 2016

It is easy to check that the logarithmic derivative has the following properties: $L(p) = \frac{1}{p}\ (p\ \text{prime});\ \ \ L(a\cdot b) = L(a) + L(b).$ Therefore if a number has prime factor decomposition $n = \prod p_i$ , then $L(n) = \sum 1/p_i$ . Now we look for two numbers, with prime decompositions $a = \prod p_i,\ b = \prod q_j$ such that $\sum \frac1{p_i} = \sum \frac1{q_j}.$ If $a$ and $b$ have a common factor, we can simply divide it out to arrive at a smaller solution; therefore we may assume that the $p_i$ and $q_j$ are entirely different. The only way in which sums of reciprocals of coprime $p_i$ and $q_j$ can be equal is if each adds up to a who number, i.e. if the exponent of each $p$ is a multiple of $p$ , etc. This leads us to the solutions $a = 2^2 = 4,\ \ b = 3^3 = 27,\ \ a + b = \boxed{31}.$ Indeed, $L(4) = \frac2 2 = 1;\ \ L(27) = \frac3 3 = 1.$

Logarithmic Derivatives of Numbers

The answer is 31.

1 solution

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