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

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In summary, "Maximizing ab with 100 Good Numbers" is a mathematical problem presented as a "Problem of the Week" on December 29, 2015. It involves finding the maximum possible value of the product of two numbers, a and b, where a and b are chosen from a set of 100 "good numbers". The "good numbers" in this problem refer to positive integers that have a prime factorization with exactly three prime factors. There is no specific set of 100 "good numbers" that need to be used for this problem. The approach for solving this problem involves finding the largest possible values for a and b while still using "good numbers". The maximum possible value for the product of a and b
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anemone
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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 $ab$-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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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 $ab$.)
 

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

What is "Maximizing ab with 100 Good Numbers"?

"Maximizing ab with 100 Good Numbers" is a mathematical problem presented as a "Problem of the Week" (POTW) on December 29, 2015. It involves finding the maximum possible value of the product of two numbers, a and b, where a and b are chosen from a set of 100 "good numbers".

What are "good numbers" in this problem?

In this problem, "good numbers" refer to positive integers that have a prime factorization with exactly three prime factors. For example, 12 is a good number because its prime factorization is 2 x 2 x 3, which has exactly three prime factors (2, 2, and 3).

Is there a specific set of 100 "good numbers" that need to be used?

No, there is no specific set of "good numbers" that need to be used for this problem. As long as the set contains 100 different "good numbers", it is valid for the problem.

What is the approach for solving this problem?

The approach for solving this problem involves finding the largest possible values for a and b, while still satisfying the condition of using "good numbers". This can be achieved by first finding the largest "good number" and then pairing it with the next largest "good number" to form a and b. This process is repeated until all pairs have been exhausted.

What is the maximum possible value for the product of a and b in this problem?

The maximum possible value for the product of a and b in this problem is 2^98 x 3^1. This can be achieved by pairing the largest "good number" (2^98) with the second largest "good number" (3^1).

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