Mathematical Thinking I, II, .... (previously Puzzle: Marrie

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Sayumi
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby Sayumi » April 14th, 2009, 7:55 pm

Nope, not the minimum number of students  :-\... though it meets the other conditions.
"It is one of those instances where the reasoner can produce an effect which seems remarkable to his neighbor, because the latter has missed the one little point which is the basis of the deduction."
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Married?)

Postby S.H. » April 14th, 2009, 9:32 pm

[spoiler]I really missed these riddles :)
301
As for the solution...
To make it simple, there are only 2 last digits that will have an excess of 1 for the multiples of 5, which is 6 and 1. But since 6 is a multiple of 2, we have to eliminate it.  For the 7 to have a last digit of 1 it needs to be multiplied by 10x + 3 (where zero is a non negative integer. then you'll have 21,91,161,231,301,...... then eliminate the impossible  :D[/spoiler]
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby Sayumi » April 15th, 2009, 5:19 am

Yay, correct answer  :D!

My way of solving it was....
[spoiler]Find the lowest common multiple of 2,3,4 and 5. That's 60. Now try which multiple of 60 plus one is a multiple of 7
61, 121, 181, 241, 301
But your way works just as well  ;D[/spoiler]
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby Holmes » April 15th, 2009, 12:10 pm

This could be a General Mathematical Thinking Thread, cause there to many problems.
Leaving the Lateral Thinking Threads for a while ...

Here´s mine:

You've been asked to buy 100 squeegies, using 100 dollars to do so. You may buy no more or less than 100 squeegies, and the total price must be exactly 100 dollars. There is no sales tax. Red squeegies cost $6.00. Yellow squeegies cost $3.00. Blue squeegies cost $0.10. How many of each must you buy?
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Married?)

Postby bash7353 » April 16th, 2009, 4:24 am

[spoiler]It works if you buy 70 blue, 29 yellow and 1 red ones.
Blue: 70*0.1=7$
Yellow: 29*3=87$
Red: 1*6=6$

7$+87$+6§=100 dollars
70s+29s+1s=100 squeegies[/spoiler]


Edit:
That's how I did it:
[spoiler]Two conditions were required in the riddle, so I put the count of each color in a variable (red -> r; yellow -> y; blue -> b) in order to formulate equations:
1) The sum of all squeegies must be 100 -> r+y+b=100; -> r=100-y-b
2) The cost must also be exactly 100 -> 6r+3y+0.1b=100

I solved the first equation for r, so that I can now plug it in in the second one:
6(100-y-b)+3y+0.1b=100

Solving that for y gives me:
y=-(59/30)b+(500/3)

After exchanging the letter b with x I plotted that function and had the software create a lookup table, so I could look for natural numbers.
When I determined the third number (how many red squeegies are required in that case) I eliminated every solution that contained a negative number (since while you can buy 5 squeegies, you can't buy -5 ones). If I didn't miss one, the solution I stated above is the only one.[/spoiler]


Edit2: If someone were to tell me my answer is correect, don't wait for me to post the next riddle, 'cause I got none.
Last edited by bash7353 on April 17th, 2009, 4:43 am, edited 1 time in total.
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby akai-kun » April 19th, 2009, 6:19 am

googleearth wrote:[spoiler]It works if you buy 70 blue, 29 yellow and 1 red ones.
Blue: 70*0.1=7$
Yellow: 29*3=87$
Red: 1*6=6$

7$+87$+6§=100 dollars
70s+29s+1s=100 squeegies[/spoiler]


Edit:
That's how I did it:
[spoiler]Two conditions were required in the riddle, so I put the count of each color in a variable (red -> r; yellow -> y; blue -> b) in order to formulate equations:
1) The sum of all squeegies must be 100 -> r+y+b=100; -> r=100-y-b
2) The cost must also be exactly 100 -> 6r+3y+0.1b=100

I solved the first equation for r, so that I can now plug it in in the second one:
6(100-y-b)+3y+0.1b=100

Solving that for y gives me:
y=-(59/30)b+(500/3)

After exchanging the letter b with x I plotted that function and had the software create a lookup table, so I could look for natural numbers.
When I determined the third number (how many red squeegies are required in that case) I eliminated every solution that contained a negative number (since while you can buy 5 squeegies, you can't buy -5 ones). If I didn't miss one, the solution I stated above is the only one.[/spoiler]


Edit2: If someone were to tell me my answer is correect, don't wait for me to post the next riddle, 'cause I got none.


Since your solution worked, I would say that your answer is correct! And because of no response from Holmes by now, I think a new quiz can start!
So, you don't find one, googleearth?
Well, then I post the next one if it's okay for you all...

So, here it is:

"You've got 60 square concrete slabs. And with this 60 slabs you shall make a rectangle with a minimal circumference. Can you tell what side-lengths the rectangle have? Explain your answer!"

(Hope you get the quiz... ;) )
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby GG.Wicken » April 19th, 2009, 7:18 am

@akai-kun:

[spoiler]Because the rectangle we are finding is created from 60 squares, hence its area is 60.

Set the square's edges length = x
Hence the rectangle's edge length = Ax and Bx
Since the squares can't be divided into pieces, A and B must be natural numbers

The rectangle's circumference: 2(A + B)
The rectangle's area: A*B = 60  -->  A = 60/B
(I removed the "x" part since it only serves as the length's unit)

Since A and B are natural numbers, B must be either 1, 2, 3, 4, 5 or 6; A will be 60, 30, 20, 15, 12 and 10 respectively
Since we want the minimum rectangle's circumference, B must be 6  -->  A is 10

The minimum circumference is 2(6 + 10) = 32 (unit is x)[/spoiler]

Right, Shuu? :D
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby Holmes » April 19th, 2009, 9:26 am

Excucez-moi pour my non reply, my Internet was dead but now it returned!

@googleearth, answer, correct!
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby akai-kun » April 19th, 2009, 12:20 pm

GG.Wicken wrote:@akai-kun:

[spoiler]Because the rectangle we are finding is created from 60 squares, hence its area is 60.

Set the square's edges length = x
Hence the rectangle's edge length = Ax and Bx
Since the squares can't be divided into pieces, A and B must be natural numbers

The rectangle's circumference: 2(A + B)
The rectangle's area: A*B = 60  -->  A = 60/B
(I removed the "x" part since it only serves as the length's unit)

Since A and B are natural numbers, B must be either 1, 2, 3, 4, 5 or 6; A will be 60, 30, 20, 15, 12 and 10 respectively
Since we want the minimum rectangle's circumference, B must be 6  -->  A is 10

The minimum circumference is 2(6 + 10) = 32 (unit is x)[/spoiler]

Right, Shuu? :D


Yeah... Correct answer!
Now, it's your turn! ;)
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GG.Wicken
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby GG.Wicken » April 19th, 2009, 1:15 pm

My turn, say? Couldn't think of any good one now, but here's one I remember the most :D

A woman sold a number of egg.
In her first sale, she sold half of the eggs AND half an egg.
In her second sale, she sold half of the remaining eggs AND half an egg.
In her third sale, she sold half of the remaining eggs AND half an egg.
After her third sale, no egg remained.

Q: What is the number of egg before selling?


P/S: If my mind's right this is 7th grade math...I struggled solving it when I was in grade 7 though :D...And yeah this kind of math can drag on to the 4th, 5th, etc. sales...
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Married?)

Postby SilverBullet » April 19th, 2009, 3:05 pm

@ GG.Wicken
[spoiler]
7 eggs.

Let E be the initial number of eggs:
First sale: E*0.5-0.5=a
Second sale: a*0.5-0.5=b
Third sale: b*0.5-0.5=0

After solving the equations you get that she had 1 egg after sell 2, 3 after the first one, and 7 at the begining.
[/spoiler]

Right, right?  ;)
Last edited by SilverBullet on April 19th, 2009, 3:09 pm, edited 1 time in total.
GG.Wicken
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby GG.Wicken » April 19th, 2009, 4:18 pm

@ SilverBullet

Yeah that's right
It's not a difficult question though :D

Your turn, I guess?
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby SilverBullet » April 19th, 2009, 5:02 pm

Ok, here's mine:

Two mathematitians met in the street after a long time without seeing each other.
- Long time no see!
- Yeah, it seems like it was yesterday...
- So, you got married, rigth?
- Yes, and I have 3 beautiful daughters.
- How old are they?
- I'm not telling you their ages, but I'm telling you that the product of their ages is 36 and their sum is 13.
The friend started making calculations and after some time said:
- I need more information
- Sure, the oldest one plays piano.
Rigth after hearing this, the friend said the answer.

How old are the daughters? Explain your answer.


I'm leaving the answer as spoiler, just in case I'm away for a long time (don't look before posting yours  ;)):
[spoiler]2, 2, 9[/spoiler]
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Married?)

Postby S.H. » April 20th, 2009, 2:40 am

[spoiler]There are 2 answers... 9 2 2 and 6 6 1...But since there is an oldest one...The answer is 9 2 2....[/spoiler]
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Re: Mathematical Thinking I, II, .... (previously Puzzle: Ma

Postby akai-kun » April 20th, 2009, 1:19 pm

S.H. wrote:[spoiler]There are 2 answers... 9 2 2 and 6 6 1...But since there is an oldest one...The answer is 9 2 2....[/spoiler]


Here's your answer:

SilverBullet wrote:I'm leaving the answer as spoiler, just in case I'm away for a long time (don't look before posting yours  ;)):
[spoiler]2, 2, 9[/spoiler]


See, correct! You can start with a new one!  ;)
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