Newspaper puzzle

The original thread is now closed apparently.
Anyway, this weeks puzzle is quite easy.Just one formula but it may cause a few moments of amusement :-
A 1x1 square grid can be made using 4 matchsticks
A 2x2 square grid can be made using 12 matchsticks.
How many matchsticks do you need to make a 700x700 square grid?
The question is always something to do with 700!!

@venda said
The original thread is now closed apparently.
Anyway, this weeks puzzle is quite easy.Just one formula but it may cause a few moments of amusement :-
A 1x1 square grid can be made using 4 matchsticks
A 2x2 square grid can be made using 12 matchsticks.
How many matchsticks do you need to make a 700x700 square grid?
The question is always something to do with 700!!

For an square grid of size "n x n" we need

4*n + 2*n*(n-1)

4*700 + 2*700*699 = 981,400 match sticks

It seems the puzzler has a bit of an obsession!

@joe-shmo said
For an square grid of size "n x n" we need

4*n + 2*n*(n-1)

4*700 + 2*700*699 = 981,400 match sticks

It seems the puzzler has a bit of an obsession!

Correct on both counts!!
The weekly puzzles column(there's "wordy" ones as well) is entitled "Mindset -by 700.
There was even one once where they printed a string of 1's and zeroes(obviously binary)and asked what they meant!!
I mean -really?
I'm sure he could have used base 7 or something but there you go.

I think this weeks puzzle is only solvable by "trial and error"
See what you think:-
The thumb wrestling championships(what!!?)divides entrants into a professional section and an amateur section.Within each section every player plays each other player exactly once.There are 700(surprise surprise!!)matches in total.How many entrants are there?
As far as I can see this reduces down to this equation:-
x^2-x +y^2-y=1400.
This is unsolvable except by trial and error.
Is there a different approach?

@venda said
I think this weeks puzzle is only solvable by "trial and error"
See what you think:-
The thumb wrestling championships(what!!?)divides entrants into a professional section and an amateur section.Within each section every player plays each other player exactly once.There are 700(surprise surprise!!)matches in total.How many entrants are there?
As far as I can see this ...[text shortened]... -
x^2-x +y^2-y=1400.
This is unsolvable except by trial and error.
Is there a different approach?

I used trial and error and a spreadsheet


Let the number of professionals be "P", let the number of amateurs be "A"

C ( P , 2 ) + C ( A , 2 ) = 700

C ( 15 , 2 ) + C ( 35,2 ) = 700

105 + 595 = 700

So:

P + A = 50

Where does your equation come from? The two can be combined to at least reduce the search to a single variable, I'm just having trouble seeing to path to your result?

Actually, your equation alone can be reduced to a single parameter search by competing the square.

P²- P = 1400 - A*(A-1)

add 1/4 to both sides:

P²- P + 1/4 = 1400 - A*(A-1) + 1/4

The LHS now factors:

( P - 1/2 )² = 1400 - A*(A-1) + 1/4

P = ( 1400 - A*(A-1) + 1/4 )^( ½ ) + 1/2

So all we need to do is find solutions to P in the natural numbers with "A" also being restricted to the natural numbers.

@joe-shmo said
I used trial and error and a spreadsheet


Let the number of professionals be "P", let the number of amateurs be "A"

C ( P , 2 ) + C ( A , 2 ) = 700

C ( 15 , 2 ) + C ( 35,2 ) = 700

105 + 595 = 700

So:

P + A = 50

Where does your equation come from? The two can be combined to at least reduce the search to a single variable, I'm just having trouble seeing to path to your result?

Obviously, you can't play yourself,so if there are x players with two players competing every match is x(x-1)/2 and similarly for y players in the other league.
Therefor you need to find out when x(x-1)/2 + y(y-1)/2 =700., which multiplied out this comes to (x squared -x) + (y squared -y )= 1400.
I don't really see how the equation with two unknowns can be combined.
I can see where your equations work but not how you find the correct numbers 15 and 35 without using trial and error .
I also used a spreadsheet to "plug in " the numbers until they add up to 700.
I started with 25 and 25 and carried on from there.
I prefer problems with a purely mathematical method.
Back to the "dots" later!!

@joe-shmo said
Actually, your equation alone can be reduced to a single parameter search by competing the square.

P²- P = 1400 - A*(A-1)

add 1/4 to both sides:

P²- P + 1/4 = 1400 - A*(A-1) + 1/4

The LHS now factors:

( P - 1/2 )² = 1400 - A*(A-1) + 1/4

P = ( 1400 - A*(A-1) + 1/4 )^( ½ ) + 1/2

So all we need to do is find solutions to P in the natural numbers with "A" also being restricted to the natural numbers.

I vaguely remember "completing the square" from quadratic equation days.
The most interesting thing about this is how you make the little 2 to represent squared!!
My computer won't do that I don't think!!

@venda said
Obviously, you can't play yourself,so if there are x players with two players competing every match is x(x-1)/2 and similarly for y players in the other league.
Therefor you need to find out when x(x-1)/2 + y(y-1)/2 =700., which multiplied out this comes to (x squared -x) + (y squared -y )= 1400.
I don't really see how the equation with two unknowns can be combined.
I can se ...[text shortened]... ied on from there.
I prefer problems with a purely mathematical method.
Back to the "dots" later!!

They turn out to be the exact same equations as mine...I didn't see that! The number of ways to select 2 people to play from "x" people ( everybody playing everyone else exactly once ) :

C ( x , 2 ) = x!/(2! * ( x-2 )! ) = x*(x-1)* (x-2)!/( 2! * ( x-2 )! ) = x*(x-1)/2

I didn't occur to me to try to reduce C( x, 2) like that...I just left it in its base form and started plugging in values to the spreadsheet making a 2D array.

@venda said
I vaguely remember "completing the square" from quadratic equation days.
The most interesting thing about this is how you make the little 2 to represent squared!!
My computer won't do that I don't think!!

Those are called "alt codes"

I get them from the following site.

https://www.alt-codes.net/

You just click on them and they automatically copy.

Then just do: Ctrl + V or Right click - paste

in the body of the text.

@joe-shmo said
They turn out to be the exact same equations as mine...I didn't see that! The number of ways to select 2 people to play from "x" people ( everybody playing everyone else exactly once ) :

C ( x , 2 ) = x!/(2! * ( x-2 )! ) = x*(x-1)* (x-2)!/( 2! * ( x-2 )! ) = x*(x-1)/2

I didn't occur to me to try to reduce C( x, 2) like that...I just left it in its base form and started plugging in values to the spreadsheet making a 2D array.

I've never actually used the array function in a spreadsheet.
I might give it a go sometime.
Thanks for the codes website info..That could be useful.

@venda said
I've never actually used the array function in a spreadsheet.
I might give it a go sometime.
Thanks for the codes website info..That could be useful.

Array formulas are definitely a handy tool to have! You can find tutorials on how they work all over the web.

@joe-shmo said
Array formulas are definitely a handy tool to have! You can find tutorials on how they work all over the web.

I've a fair idea of how they work,I just haven't found a use for them yet.
Still not looked at the "dots".Been a bit busy with other stuff.I'll get round to them eventually I expect.

This weeks puzzle is not quite as easy as it first appears.
I'm still working on the formula for all cases and I haven't looked at the answer.
Here's the puzzle.No mention of 700 for a change although that could be part of the answer:-
With a deck of 52 cards you can build a triangular house of cards five storeys high,but would have 12 cards left over.What is the fewest(no zero) number of decks you would need if you wanted to use all the cards to build a singular triangular house with no cards left over?

@venda said
This weeks puzzle is not quite as easy as it first appears.
I'm still working on the formula for all cases and I haven't looked at the answer.
Here's the puzzle.No mention of 700 for a change although that could be part of the answer:-
With a deck of 52 cards you can build a triangular house of cards five storeys high,but would have 12 cards left over.What is the fewest(n ...[text shortened]... if you wanted to use all the cards to build a singular triangular house with no cards left over?

Reveal Hidden Content
Reveal Hidden Content
Reveal Hidden Content