Newspaper puzzle

@venda

I wouldn't think this one can be solved logically, but Joe may show differently...

@bigdoggproblem said
@venda

I wouldn't think this one can be solved logically, but Joe may show differently...

@joe-shmo said
It can be solved analytically, but I don't believe exactly ( unless there is some number theory does the last step ). What I mean is that we can write a formula for the number of cards in an "n" story triangular card house, but we have to do a parameter search to find the solutions that exist with "n" as a natural number.

@venda said
Been a bit busy with other stuff so I haven't looked at the problem again or your hidden answers yet.
Do you have the actual answer,or do I need to look in the paper to confirm?

That's right Joe
Well done

Let "n" be the number of stories, and "c" be the number of cards to construct it. Let the cards used to form the side walls be "d", and floor above them be "f"

n = 1

/\

c = d = 2

n = 2

../\
..__
/\/\

c = 2*d + f +[ d ]

n = 3

../\
..__
./\/\
.__ __
/\/\/\

c = 3d + 2f + [ 2d + f + d ]

n= 4

c = 4d+3f + [3d + 2f + 2d + f + d ]

For a house of "n" stories

c = nd + (n-1)f + (n-1)d + (n-2)f + (n-2)d +...+ 2f + 2d + f + d

rearrange the terms grouping by type


c (n) = { 1 + 2 + ... + (n-2) + ( n-1 ) + n }d + { 1 + 2 + ... + (n-2) + ( n-1 ) }f

Sum of the first "n" natural numbers for the number of sidewall units:

1 + 2 + ... + (n-2) + ( n-1 ) + n = n *( n + 1 )/2


Sum of the first "n-1" natural numbers for the number of floor units:

1 + 2 + ... + (n-2) + ( n-1 ) = ( n-1 )* n /2

So :

c(n) = d* n * ( n + 1 )/2 + f* ( n-1 ) *n /2

where d = 2 and f = 1

this simplifies algebraically to:

c(n) = n*( 3*n+1 )/2

From here I just used the quadratic formula to solve for "n" sweeping through "k"

n*( 3*n+1 )/2 = 52*k

first( non-zero) solution:

n = 13 , k = 5

I guess my question to people better at number theory:

can we find the solutions for "k" from:

k = n*( 3n+1 )/104

the prime factorization of 104 = 2*2*2*13

The numerator has to be even to be evenly divisible by 104.

n = even → 3n+1 = odd
n= odd → 3n+1 = even

I never no where to go with this type of question!

@joe-shmo said
I guess my question to people better at number theory:

can we find the solutions for "k" from:

k = n*( 3n+1 )/104

the prime factorization of 104 = 2*2*2*13

The numerator has to be even to be evenly divisible by 104.

n = even → 3n+1 = odd
n= odd → 3n+1 = even

I never no where to go with this type of question!

I don't really understand this weeks puzzle,I'm not that familiar with pyramid geometry but I'm sure some of you will be able to solve it:-
A Faberge egg collection may only be displayed when arranged as triangular pyramids of 1,4,10,20,35 and so on.
700(as usual!!) eggs are on display.What is the smallest number of pyramids that can make up the display ?

@venda said
I don't really understand this weeks puzzle,I'm not that familiar with pyramid geometry but I'm sure some of you will be able to solve it:-
A Faberge egg collection may only be displayed when arranged as triangular pyramids of 1,4,10,20,35 and so on.
700(as usual!!) eggs are on display.What is the smallest number of pyramids that can make up the display ?

Reveal Hidden Content

Reveal Hidden Content

Reveal Hidden Content

@joe-shmo said
Hey venda, If you want to give it a go the author is telling you how many eggs are used to construct a triangular pyramid ( so we don't really have to know how to build them per se) .

There are triangular pyramids of 1 egg., 4 eggs, 10 eggs, 20 eggs, 35 eggs, etc... The puzzle is to figure out the pattern for the sequence above ( no geometry necessary ).

Hint:
[hidd ...[text shortened]... 120 + 165 + 220 = 700 Just have to come up with the right question to ask to suss it out! [/hidden]

@venda said
Re another thing-The question asks the smallest number of pyramids.
Your answer is correct.

This weeks puzzle should be good for all you mathematicians who are far more educated than me:-
An amplifier has 100 dials that go up to 11.Turning them past 11 returns them to zero.The first roadie turns every dial to 1.The second roadie turns dials 2,4,6 .... up by one.the third roadie turns dials 3.6.9....up by one and so on.After the hundredth roadie finishes which dials are on zero ? . Which dials would be on zero if they only went up to 10?
Can't imagine a band having a hundred roadies but there you go!!

@venda said
This weeks puzzle should be good for all you mathematicians who are far more educated than me:-
An amplifier has 100 dials that go up to 11.Turning them past 11 returns them to zero.The first roadie turns every dial to 1.The second roadie turns dials 2,4,6 .... up by one.the third roadie turns dials 3.6.9....up by one and so on.After the hundredth roadie finishes which dial ...[text shortened]... o if they only went up to 10?
Can't imagine a band having a hundred roadies but there you go!!

Reveal Hidden Content

@joe-shmo said
Im not aware of methods for counting factors of numbers analytically, but I'm sure they exist.