A few questions

bobbym · 2012-12-03 17:27:50

Hi Avon;

bobbym, I apologise if I have caused any offence

No need for an apology I was not offended, I just wanted Agnishom to know I was not annoyed or put off that he was asking questions about the alleged proof. As you stated he was correct is asking for more.

I think you are missing something.

Hmmm, since the day I was born that has been true. You are correct, the hole in my reasoning is in 2). I thought I could get around that and thus leave out the derivation as he supplied in the book.

anonimnystefy · 2012-12-03 21:58:47

bobbym wrote:

Hi;
Call there ages a,b,c,d.
There are four girls needed to make 6 pairs.

Hi bobbym

That isn't the only solution.

Last edited by anonimnystefy (2012-12-03 21:59:17)

bobbym · 2012-12-03 22:04:19

Hi;

Are you sure about that?

anonimnystefy · 2012-12-03 22:06:25

Yes.

Did ypu try permutating the rhs' of those equations?

Last edited by anonimnystefy (2012-12-03 22:07:47)

bobbym · 2012-12-03 22:13:32

Yes, but that would only change the order. Since the assignments of a,b,c,d is arbitrary the order does not count. No matter how we permute it the 4 girls ages are the same.

phrontister · 2012-12-03 23:30:05

Hi all,

Here's what I did...

There are only 6 possible pairings, as follows:

a + b = 30
a + c = 33
a + d = 41
b + c = 58
b + d = 66
c + d = 69

From this, 3a + 3b + 3c + 3d = 297

Substituting (to leave one variable...for which I chose d):

123 - 3d + 198 - 3d + 207 - 36 + 3d = 297
6d = 231
d = 38.5

Substituting that result to find the other ages:
a = 41 - d = 41 - 38.5 = 2.5
b = 66 - d = 66 - 38.5 = 27.5
c = 69 - d = 69 - 38.5 = 30.5

It seems to me that they're the only solutions.

bobbym · 2012-12-04 00:08:52

Hi phrontister;

Did you see my post #24?

Agnishom · 2012-12-04 00:27:31

Hmm... I also tried doing it on my own
But I was confused.
However, Thank You!

phrontister · 2012-12-04 00:28:16

Hi Bobby,

Yes, I did, but as I didn't know how you got to your answer I thought I'd give it a try with the little I know.

I suppose there's some other way of doing this that conforms to the other posts in the thread...which I didn't follow, as I don't think I've done this type of problem before.

bobbym · 2012-12-04 00:30:48

Hi;

I am super busy with another problem so I have not had time to check. Does your solution agree with mine?

phrontister · 2012-12-04 00:32:00

Yes, exactly the same. The only difference is that yours are fractions and mine are decimal.

bobbym · 2012-12-04 00:40:46

Okay, thanks for checking. I think Mr. Agnishom's question has been answered.

Agnishom · 2012-12-04 00:45:57

My answers match too....

This one is very basic but I am not aware of the basic concepts used
Question 4:
Find the remainder when 21990 is divided by 1990.

phrontister · 2012-12-04 00:49:14

Hi Bobby,

Yes, I thought you probably answered it but I don't know enough about this topic to know that for sure. I was just throwing my tuppence-worth into the ring in case it might mean something to someone.

bobbym · 2012-12-04 00:54:03

Hi phrontister;

I do not mind at all.

Hi Agnishom;

That is a modulo operation.

Agnishom · 2012-12-04 00:55:01

Yes I know it is called a modulo operation, .... How do you tackle it?

bobbym · 2012-12-04 01:02:30

When they are both positive like that the simplest way is

take the integer part of the answer which is 11.

We say:

phrontister · 2012-12-04 01:03:39

Here's how I think about it:

I'll let bobbym explain modulo. I know what it does, but I forget why.

bobbym · 2012-12-04 01:06:34

Hi phrontister;

Sometimes when the numbers are small successive subtraction is fastest. Of course Mathematica has a built in function.

Agnishom · 2012-12-04 01:07:47

I am terribly embarassed about it

I meant to say:

Find the remainder when

is divided by 1990

phrontister · 2012-12-04 01:17:24

Hi Bobby,

I've only comparatively recently started thinking 'floor function' instead of 'integer'...and I also only use the dot instead of the multiplication sign rarely.

My instinct was to write the equation in my previous post as

phrontister · 2012-12-04 01:28:26

Hi Agnishom,

Actually, from what I see of how you typed that into your post, you meant this:

Find the remainder when

is divided by 1990.

But I'm off to bed now (very early start in the morning), so I'll leave the answer to others.

Agnishom · 2012-12-04 01:30:53

Yes sorry Please don't mind
I have some problems with tackling that latex

scientia · 2012-12-04 04:41:52

Last edited by scientia (2012-12-04 04:44:11)

Agnishom · 2012-12-04 14:23:41

scientia wrote:

We also know that
since

Will you please explain this line?

Math Is Fun Forum

#26 2012-12-03 17:27:50

Re: A few questions

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Re: A few questions

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