MHB How Can You Maximize the Product of Two Numbers with 100 Good Numbers?

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    2015
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To maximize the product of two numbers \(a\) and \(b\) given that there are exactly 100 good integers \(k\) satisfying the triangle inequality with sides \(\log a\), \(\log b\), and \(\log k\), one must analyze the conditions under which these integers are defined. The triangle inequality leads to constraints on the logarithmic values, which can be translated into conditions on \(a\) and \(b\). By optimizing these values while adhering to the constraints, the maximum product \(ab\) can be achieved. The solution provided by kaliprasad successfully identifies the optimal values for \(a\) and \(b\) that yield this maximum product. This problem illustrates the interplay between logarithmic functions and geometric properties in maximizing products under specific conditions.
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Here is this week's POTW:

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A man chooses two positive integers $a$ and $b$. He then defines a positive integer $k$ to be good if a triangle with side lengths $\log a$, $\log b$ and $\log k$ exists. He finds that there are exactly $100$ good numbers. Find the maximum possible value of $ab$-----

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Congratulations to kaliprasad for his correct solution::)

Solution from kaliprasad:

Without loss of generality we can assume $a\ge b$.

As $\log\, a$, $\log\, b$ and $\log\, k$ form a triangle we have $\dfrac{a}{b}< k < ab$.

So maximum value of k is ab-1 and minimum $\lfloor\dfrac{a}{b}+1\rfloor$.

As there are 100 values of k we have:

$ab-1 - \lfloor\frac{a}{b}+1\rfloor = 99$

or $ab - \lfloor\frac{a}{b}\rfloor= 101$

For the same $a$, as $b$ increases $\frac{a}{b}$ decreases and so $ab$ decreases and we expect it to be larger when $b$ is in minimum.

$b\ge 2$ (as it is integer ) else $\log\, b$ is undefined.

Put $b = 2$ to get $a = 67$ and $ab= 134$ is maximum.

(Note $b= 5$ and $a = 21$ and $ab = 105$ is minimum and these are only 2 solutions for $ab$.)
 
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