which number is larger?

Oct 2009
273
2
Fellas, I have a question regarding quantitative comparisons.

Say we have, in column A, (78)(243) and in column B, (77)(244).

I'm asked to determine which is larger.

Is it a good strategy to pick out the first two numbers in each variable? So, in column A we would take (78)(24) and B would be (77)(24). Thus A is larger because 78>77.

What about A= (90,021)(100,210) and B=(90,210)(100,021)? Using my method, we have (900)(100) in A and (902)(100) in B. Thus, B is larger.

Is it possible to do this? Is there some law?
 

Ackbeet

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Jun 2010
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What does the notation (78)(243) mean? I mean, I normally think usual arithmetic product of the two numbers. But in number theory, you might have a different context.
 

Ackbeet

MHF Hall of Honor
Jun 2010
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Ok, so I would say that, given two numbers whose sum is a constant, their product is maximized when the two numbers are closer together (global max is when they are equal). So I would say instantly that the product in Column A is greater. You can prove this using calculus: assume \(\displaystyle x+y=\text{constant}=c.\) The goal is to maximize the product \(\displaystyle xy=x(c-x).\) So, let

\(\displaystyle f(x)=x(c-x).\) Then

\(\displaystyle f'(x)=c-2x.\) Setting this equal to zero implies

\(\displaystyle c-2x=0,\) or \(\displaystyle x=c/2=y.\) Done.
 

undefined

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Mar 2010
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Fellas, I have a question regarding quantitative comparisons.

Say we have, in column A, (78)(243) and in column B, (77)(244).

I'm asked to determine which is larger.

Is it a good strategy to pick out the first two numbers in each variable? So, in column A we would take (78)(24) and B would be (77)(24). Thus A is larger because 78>77.

What about A= (90,021)(100,210) and B=(90,210)(100,021)? Using my method, we have (900)(100) in A and (902)(100) in B. Thus, B is larger.

Is it possible to do this? Is there some law?
(78)(243) = (77 + 1)(243) = 77 * 243 + 243

(77)(244) = (77)(243 + 1) = 77 * 243 + 77
 
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Ackbeet

MHF Hall of Honor
Jun 2010
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CT, USA
You're welcome. undefined and AsZ's methods are also both entirely valid. AsZ's method might be generalized by proving that

ab > (a-m)(b+m).