Highest Product: 7.5 x7.5=56.27
I can't really think of a way to prove this right now (except for a way which would involve Pythagoras and derivatives.... and I can't be bothered to formulate that!)
but it can be explained using approximation.
0x15= 0
1x14= 14
2X13= 26
...
6x9= 54
7X8= 56
8X7= 56
9X6= 54
...
from here we could continue this with 7.9x7.1=56.09 ...
And would eventually end up with 7.5x7.5
The result won't change if we can use negative numbers as well. When the sum is +15 and one of numbers is negative the other one has to be positive. When multiplying a positive and a negative number the result will always be negative and therefore lower than 56.25.
Now... another maths puzzle....
John is celebrating his 36th birthday. One of the guests asks him how old his brother Joe is.
John answers: I am now double as old, as Joe was, when I was as old, as Joe is now.
How old is Joe? (assuming that John's statement is correct of course)
Spoiler:
how would you solve it the Pythagoras way
edit: the only way i saw was the calculus way, by making it an optimization problem. Taking the derivative of the equation when its all put together and finding the max value
ie
x+y=15
z=x*y
then put into terms of one indep variable and then differentiate in respect to to your independent variable
i havent done one of these problems in a loooooong time so forgive me if part of its rusty if it is
Last edited by allstar1234 on April 13th, 2009, 10:58 am, edited 1 time in total.
Sayumi wrote:
John is celebrating his 36th birthday. One of the guests asks him how old his brother Joe is.
John answers: I am now double as old, as Joe was, when I was as old, as Joe is now.
How old is Joe? (assuming that John's statement is correct of course)
Highest Product: 7.5 x7.5=56.27
I can't really think of a way to prove this right now (except for a way which would involve Pythagoras and derivatives.... and I can't be bothered to formulate that!)
but it can be explained using approximation.
0x15= 0
1x14= 14
2X13= 26
...
6x9= 54
7X8= 56
8X7= 56
9X6= 54
...
from here we could continue this with 7.9x7.1=56.09 ...
And would eventually end up with 7.5x7.5
The result won't change if we can use negative numbers as well. When the sum is +15 and one of numbers is negative the other one has to be positive. When multiplying a positive and a negative number the result will always be negative and therefore lower than 56.25.
Now... another maths puzzle....
John is celebrating his 36th birthday. One of the guests asks him how old his brother Joe is.
John answers: I am now double as old, as Joe was, when I was as old, as Joe is now.
How old is Joe? (assuming that John's statement is correct of course)
Spoiler:
how would you solve it the Pythagoras way
edit: the only way i saw was the calculus way, by making it an optimization problem. Taking the derivative of the equation when its all put together and finding the max value
ie
x+y=15
z=x*y
then put into terms of one indep variable and then differentiate in respect to to your independent variable
i havent done one of these problems in a loooooong time so forgive me if part of its rusty if it is
at first: @ Sayumi: correct!
Spoiler:
(56.25)
--> no problem, only approxymation is OK...
second: @ allstar1234: also, a good way...
third: @ all: I think a rule for the game should be there:
"Post your answers only in SPOILERS and wait for the resolution of the newest quiz from the quizmaster before you start a new quiz!"
Would be great if we all could stick to this rule because there can be false answers...
Thx.
He's named Cool Kid and they named me the agent who feared them most...: Shuichi Akai !
Highest Product: 7.5 x7.5=56.27
I can't really think of a way to prove this right now (except for a way which would involve Pythagoras and derivatives.... and I can't be bothered to formulate that!)
but it can be explained using approximation.
0x15= 0
1x14= 14
2X13= 26
...
6x9= 54
7X8= 56
8X7= 56
9X6= 54
...
from here we could continue this with 7.9x7.1=56.09 ...
And would eventually end up with 7.5x7.5
The result won't change if we can use negative numbers as well. When the sum is +15 and one of numbers is negative the other one has to be positive. When multiplying a positive and a negative number the result will always be negative and therefore lower than 56.25.
Now... another maths puzzle....
John is celebrating his 36th birthday. One of the guests asks him how old his brother Joe is.
John answers: I am now double as old, as Joe was, when I was as old, as Joe is now.
How old is Joe? (assuming that John's statement is correct of course)
Spoiler:
how would you solve it the Pythagoras way
edit: the only way i saw was the calculus way, by making it an optimization problem. Taking the derivative of the equation when its all put together and finding the max value
ie
x+y=15
z=x*y
then put into terms of one indep variable and then differentiate in respect to to your independent variable
i havent done one of these problems in a loooooong time so forgive me if part of its rusty if it is
Spoiler:
ummm.. yeah... I was kinda sleepy when I wrote that, so it probably doesn't make sense... sorry for that
I guess when solving it geometrically it would be more like having a rectangle which would have with the sidelengths of a and (15-a). Now calculating the maximum area for that rectangle would give us the solution-it is pretty much exactly the same thing as you suggested...
Hmm... just something I just came up with just now. When applying Pythagoras to that having the equation a²+(15-a)²=c², c would be at it's minimum when a=7.5 ... so when doing the whole thing backwards it is possible to link it to Phythagoras... somehow... in a complicated way... (I think there was something like this in our maths book last year - how the ratio of the sideslengths influences the diagonal, so that's probably what made me think of Pythagoras...)
Now for the solutions
@kat1214young: Nope, not the right answer
@cancercani: Yeah, that's it! I guess it was a bit of an English question as well, but my old maths teacher used to give us riddles like that, so I always think of them as maths questions
The next quiz is still the one kat1214young posted!
"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."
Sherlock Holmes
Suppose we are given an ordinary chessboard (8x8) and 32 dominoes (2x1).
Obviously, the dominoes can be arranged on the board to cover it completey.
Now, two opposite corner squares are removed. Determine whether or not 31
dominoes can be arranged on the reduced board to cover it exactly.
enjoy~
Spoiler:
It's not possible! When having dominoes on a chessboard they will always cover one black and one white position. When removing two squares from opposite corners, they will be of the same colour. Now there are two more squares of one colour than there are of the other one, so it's not possible to cover all of the remaining squares with dominoes.
Last edited by Sayumi on April 13th, 2009, 1:58 pm, edited 1 time in total.
"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."
Sherlock Holmes
ummm.. yeah... I was kinda sleepy when I wrote that, so it probably doesn't make sense... sorry for that
I guess when solving it geometrically it would be more like having a rectangle which would have with the sidelengths of a and (15-a). Now calculating the maximum area for that rectangle would give us the solution-it is pretty much exactly the same thing as you suggested...
Hmm... just something I just came up with just now. When applying Pythagoras to that having the equation a²+(15-a)²=c², c would be at it's minimum when a=7.5 ... so when doing the whole thing backwards it is possible to link it to Phythagoras... somehow... in a complicated way... (I think there was something like this in our maths book last year - how the ratio of the sideslengths influences the diagonal, so that's probably what made me think of Pythagoras...)
Now for the solutions
@kat1214young: Nope, not the right answer
@cancercani: Yeah, that's it! I guess it was a bit of an English question as well, but my old maths teacher used to give us riddles like that, so I always think of them as maths questions
The next quiz is still the one kat1214young posted!
Ah i see how Pythagoras can work when its done backwards.
Spoiler:
Also i just thought of something that should have been a little obvious.the maximized product for a desired sum is always half the sum squared(IE a square shape makes the most use out of an area). I did tons of these in high school during my calc classes and GEOM ,still it escaped me ha :P.Kinda had a little brain fart
Suppose we are given an ordinary chessboard (8x8) and 32 dominoes (2x1).
Obviously, the dominoes can be arranged on the board to cover it completey.
Now, two opposite corner squares are removed. Determine whether or not 31
dominoes can be arranged on the reduced board to cover it exactly.
enjoy~
Spoiler:
It's not possible! When having dominoes on a chessboard they will always cover one black and one white position. When removing two squares from opposite corners, they will be of the same colour. Now there are two more squares of one colour than there are of the other one, so it's not possible to cover all of the remaining squares with dominoes.
Yep! That's correct and also the exact explanation. Great job
Yay !-But that means i have to find another one...
Okay, got one...
A primary school goes on a trip. The teachers want to make the students line up, with the same number of students in each line. First they try pairs, but there is one student left without a partner. Then they try lines of three, but again one student is standing alone. The teachers also try lines of four, five and six, but there is always one (only one, not more) student left without a group. When forming lines with seven students in each line it works out perfectly.
What is the minimum number of students that go on the trip?
Last edited by Sayumi on April 14th, 2009, 5:26 pm, edited 1 time in total.
"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."
Sherlock Holmes
Sayumi wrote:
Yay !-But that means i have to find another one...
Okay, got one...
A primary school goes on a trip. The teachers want to make the students line up, with the same number of students in each line. First they try pairs, but there is one student left without a partner. Then they try lines of three, but again one student is standing alone. The teachers also try lines of four, five and six, but there is always one student left without a group. When forming lines with seven students in each line it works out perfectly.
What is the minimum number of students that go on the trip?
Spoiler:
49.
Last edited by chubs191 on April 14th, 2009, 10:09 pm, edited 1 time in total.
Sayumi wrote:
Yay !-But that means i have to find another one...
Okay, got one...
A primary school goes on a trip. The teachers want to make the students line up, with the same number of students in each line. First they try pairs, but there is one student left without a partner. Then they try lines of three, but again one student is standing alone. The teachers also try lines of four, five and six, but there is always one student left without a group. When forming lines with seven students in each line it works out perfectly.
What is the minimum number of students that go on the trip?
Spoiler:
49. If you really need an explanation I'll give you one.
Spoiler:
...is always one student left without a group...
Not, the correct answer...
49 students in lines of five would leave four students without a group, not just one!
"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."
Sherlock Holmes
Sayumi wrote:
Yay !-But that means i have to find another one...
Okay, got one...
A primary school goes on a trip. The teachers want to make the students line up, with the same number of students in each line. First they try pairs, but there is one student left without a partner. Then they try lines of three, but again one student is standing alone. The teachers also try lines of four, five and six, but there is always one student left without a group. When forming lines with seven students in each line it works out perfectly.
What is the minimum number of students that go on the trip?
Spoiler:
49. If you really need an explanation I'll give you one.
Spoiler:
...is always one student left without a group...
Not, the correct answer...
49 students in lines of five would leave four students without a group, not just one!
I get it! Thanks.
Spoiler:
2401
Last edited by chubs191 on April 14th, 2009, 7:24 pm, edited 1 time in total.
!-But that means i have to find another one...
