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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

DINAH SHORE's

*Last edited by phrontister (2010-02-10 01:41:05)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi phrontister;

Excellent!

Problem #25:

Just a little tedious:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi mathsyperson;

mathsyperson wrote:

I agree that numerical methods are usually quicker and easier to use, but I still think that it's best to be algebraic whenever possible.

It's just so much nicer to be *absolutely correct* than just very accurate. Numerical methods are fine for any and all practical purposes, but to get a closed-form expression is much more pleasing.

No way to explain my way out of this. If I say I didn't notice your reply that's worse than ignoring it. Sorry for missing it.

Me too! But I do enjoy the shock value they have on some type of people who can only deal with 1,2,3,4 and the square root of 2. Such types hate numerical solutions and use words like ugly and uninspired.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi;

Problem #26:

This one is not too difficult:

A race is 20 miles long, Anthony beats Billy by 2.5 miles and Anthony beats Charlie by 6 miles. The runners all ran at the same speed. How many miles did Billy beat Charlie by?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

The runners all ran at the same speed.

If they all ran at the same speed as each other for the whole race, then

If they each maintained their own speed for the whole race then

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi;

The phrasing is a little weird on this problem. I think it should be phrased they all ran at a constant speed through out the race.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

bobbym wrote:

The phrasing is a little weird on this problem.

Yes, I thought it was odd, ambiguous or something...

...which is why I gave 2 answers.

How did they do?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi phrontister;

I am getting a different answer. This is what I did.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Made a meal of that one! Your answer explains it well.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

Following your clear explanation of your solution to the race problem here's my attempt at explaining the equation in my last post:

*Last edited by phrontister (2010-02-16 00:13:37)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi;

I understand. Oddly enough I have trouble thinking about it like that. Is that way clearer for you?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

Is that way clearer for you?

The two ways are equally clear to me. For problems that lend themselves to that approach I prefer to go that way because I use my scientific calculator a lot and like to try to find concise methods to suit calculator entry if possible.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

This is a nice one:

Problem #27:

If a + b + c = 50

What is the maximum value of a^4*b^5*c^6.

a) What is the answer when a,b,c>0 and real.

b) What is the answer when a,b,c>0 and integers.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

If a + b + c = 50

What is the maximum value of a^4*b^5*c^6

*Last edited by phrontister (2010-02-23 12:58:36)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi phrontister;

You solved for a,b,c > 0 and a,b,c are integers. The original problem does not state that. But I liked your solution so I have modified the original question to have 2 parts. Good work!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

*Last edited by phrontister (2010-02-23 13:54:27)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi phrontister;

That is extremely close.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

That is extremely close.

Yes, I should have shown fractions. I was going to but didn't think to use improper fractions, and as MIF only provides the "½" fraction symbol I opted to display my answer with the recurring digits after the decimal point. I could've tried LaTeX, but this was easier and I didn't realise the two modes gave different results.

WolframAlpha's "decimal approximation" of the fractional input is exactly the same as the one given by my calculator for which I used decimal input...up to WolframAlpha's maximum output of 3705 digits. Maybe it can display more digits, but I didn't know how to get it to do that (my calculator displays up to 5012 digits).

I wonder what the actual difference is between the correct answer and the approximations given by both WolframAlpha and my calculator.

I used Excel to solve both parts: spreadsheet for 'b' and Solver for 'a'.

Btw, your 'hidden text' label shows "b" instead of "a".

*Last edited by phrontister (2010-02-24 01:39:09)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi;

Thanks, I am correcting the error of a and b. Yea, sometimes Wolfram is enigmatic. The package goes to any number of digits but they are stingy on their page at times. I've written them repeatedly to provide Mathematica in its entirety at alpha, but why the heck should they listen to me.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Problem #28:

Can you get the coefficient of x^28 in the expansion of ( x + 2 ) ^20 * ( x^2 - 1 )^5?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

Hi Bobby,

I spotted an error on MAlpha. After clicking "More digits" in their "Result" window the "10" disappears from the exponent reference at the end of the "Copyable plaintext" output. Not always...maybe only after a couple of clicks on "More digits", or maybe after the last-available click on it.

I've emailed them about it.

*Last edited by phrontister (2010-02-24 11:58:00)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi;

What were you entering when it happened?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,295

It happened for these three:

13.33333333333333333333^4*16.66666666666666666666^5*20^6

13.333333333333333333333333333333^4*16.666666666666666666666666666666^5*20^6

13.3333333333333333333333333333333333333333333333333333333333333333333333333333333333333

3333333^4*16.666666666666666666666666666666666666666666666666666666666666666666666666666

66666666666666666^5*20^6

I was checking the results against those of my calculator.

I notice now that the error only occurs at the last click of the "More digits" button.

*EDIT*: Looks like that long third entry wrecked the page layout and the word wrap doesn't wrap. I've split it into 3 lines.

*Last edited by phrontister (2010-02-24 21:36:25)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Hi phrontister;

It looks like yet another bug. Here is one you might like, I got it from Babai:

Problem #29

Each letter stands for a different digit. Leading zeros are not allowed.

TEN + TEN + NINE + EIGHT + THREE = FORTY

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,372

Problem #30:

Simple probability and equation solving:

A company designs 2 new bombers. One is a 3 engine bomber and the other is a 5 engine bomber (5 engines?). Each plane can make it home with a majority of its engines operating. Each engine has a probabilty p of failure. For what values of p is the 3 engine bomber more likely to return home?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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