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Magpie
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math puzzle
I recently saw on TV a math puzzle so decided to take a crack at it. It was more difficult than I thought it would be. See what you think:
Below is a crude drawing of a staircase with 4 steps. It is made using 10 blocks. The challenge is to find an algebraic formula that relates the
number of blocks, N, with the number of steps, s, for any such staircase, no matter how many steps.
See what you can do and post the results, with explanation of how you got your answer. Looking this up in a handbook will only earn you minimal
points. 
The single most important condition for a successful synthesis is good mixing - Nicodem
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pantone159
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The first step (s=1) takes one block (N(1)=1).
The second step (s=2) takes two more blocks (N(2)=N(1)+2)
The third step (s=3) takes three more blocks (N(3)=N(2)+3)
Generalizing...
N(s) = 1+2+3+...+s
This sum is equal to (1/2)*(s+1)*s, the formula is famous due to Gauss[1], you can get it by calculating the average of the terms... the average of
the first and last is (1+s)/2, and the average of the second and second-from-last is the same, and so on, and there are s terms...
So,
N(s) = (s*s + s) / 2
If I want to solve for s(N), then re-arrange this as:
s*s + s - 2*N = 0
then use the quadratic equation to get:
s = (1/2)*(-1 +/- sqrt (1 + 8N))
and only the + sign is meaningful, so the final equation for N is:
N = (1/2) * (sqrt(1 + 8N) - 1)
To test, plug in s=4, then N = (1/2) * (sqrt(81) - 1) = (1/2) * (9 - 1) = 4, which is correct.
[1] - According to legend, when Gauss was a kid, one of his teachers tried to punish him by making him sum all numbers from 1 to 100, expecting that
this tedious sum would take a long time, by the end of which the impertinent Gauss would have learned his lesson, but unfortunately for the teacher,
Gauss had the answer immediately.
[Edited on 19-5-2007 by pantone159]
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Magpie
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Yes, pantone, that is the answer I got also, but with a totally different approach. I didn't realize it was a known equation until I found it in my
CRC math handbook. I wonder if others will have different methods too.
That is an interesting story about Gauss. I'm sure that you've seen Bell's Men of Mathematics which has many interesting stories about
famous mathematicians.
The single most important condition for a successful synthesis is good mixing - Nicodem
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pantone159
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Re the formula you looked up, do you mean for 1+2+3+...+s ?
If you are interested in the formula for 1+4+9+16+...+s^2, this came up on a rival chemistry forum once:
http://www.chemicalforums.com/index.php?topic=9973.0
That one is harder, the first time I answered this I did have to look it up, I finally came up with a way of deriving it, but in truth I looked that
up too. 
I haven't seen that book about mathematicians, I'll keep my eye open. My library is actually very thin regarding history of math.
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Magpie
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| Quote: |
Re the formula you looked up, do you mean for 1+2+3+...+s ?
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Yes, that is in the CRC math handbook.
The way I solved it was based more on a geometrical interpretation as this is the way the problem was presented on TV, ie, as a staircase, as on my
drawing. I will show this derivation later but don't want to prejudice other potential solvers at this time.
The single most important condition for a successful synthesis is good mixing - Nicodem
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JohnWW
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Chapter 2 of Perry's Chemical Engineers' Handbook also has the formulae.
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woelen
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This is a special case of a general formula for summation of powers of n.
For computing integrals, we simply have ∫x^ndx = 1/n * x^(n+1)
A similar thing, however, exists for summations of x^n. ∑x^n for x = 1 to k is a power series in k, with highest power being k^(n+1). Computing
the sum of any polynomial of power n can be done as follows:
Assume that the result is a polynomial of degree n+1 with unknown coefficients. Now compute the sum for k = 1, plug in the value k = 1 into the
polynomial and make this equal to the answer. Do the same for k = 2, until k = n + 2. Now you have n+2 linear equations in the n+2 coefficients and
solve the system. There are explicit formulas for sums of this kind, but I don't remember them. I use the method of solving systems of equations. It
always works.
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MadHatter
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Basic Algebra
Using Gauss's N * (N +1) / 2, substract the number of blocks above a particular step, S, from
the number of blocks for the staircase. Assuming the bottom step is step 1, I get the following:
If N is the number of steps in the staircase then the total number of blocks is (N^2 + N) / 2.
By substituting (N - S) for N, the number of blocks above step S is (N - S) * (N - S + 1) / 2.
That opens up to (N^2 - 2NS + N + S^2 - S) / 2. Since we're subtracting it, change the signs
(-(N^2) + 2NS - N - (S^2) + S) / 2.
From the staircase total, N^2 and N cancel out leaving:
(2NS - (S^2) + S) / 2
N = 10(steps in the staircase). Calculate at S = 6(step number):
B = (2*10*6 - (6^2) + 6) / 2
B = (120 - 36 + 6) / 2
B = 90 / 2
B = 45 blocks
From opening of NCIS New Orleans - It goes a BOOM ! BOOM ! BOOM ! MUHAHAHAHAHAHAHA !
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Magpie
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I'm surprised at the many ways to solve this problem. Here's the way I did it:
The drawings below depict 3 succesive staircases with s = 4, 5, and 6 respectively. The dotted lines are there to show complementary staircases that
form squares with the original solid line staircases.
Looking at these drawings you can see that:
(1) The dotted line staircase for s is equal to the solid line staircase for s-1. Therefore, using subscripts
N(s) = s + N(s-1)
(2) the square formed for s steps has blocks equal to those of the staircase for s plus those for s-1. Therefore
N(s) = (s^2) - N(s-1).
Solving these two equations simultaneously gives
N(s)-s = (s^2) - N(s)
2N(s) = (s^2) + s
and finally
N(s) =(s^2 + s)/2. Removing the subscript gives
N = (s^2 + s)/2
[Edited on by Magpie]
[Edited on by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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franklyn
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Here's one that's not obvious
Hint : don't assume anything !
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chemoleo
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There is trickery:
Something wrong with the angles, the top green triangle (5x2) is 5.4x2 below within the red triangle.
Similarly, the green triangle at the bottom (5x2) is 5x1.8 in the red triangle above. Indeed, the lines of the hyperthenuse (term? - the diagonal) are
not straight for either of them...the 5x2 relationship doesnt hold. So it is actually a square in both cases, which looks like a triangle...almost.
There's the missing square!
Nice deception though, making up for a nice neat 1x1 square!
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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Magpie
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Yes, a clever optical illusion. The smallest angle for the green triangle is 21.8 deg whereas the smallest angle for the red triangle is 20.6 deg.
Therefore the 2 triangles are not similar. But they appear to be at first glance.
[Edited on by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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12AX7
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Sure...
Start by writing out the figures.
N(1) = 1
N(2) = 1 + 2 = 3
N(3) = 1 + 2 + 3 = 6
N(4) (shown) = 1 + 2 + 3 + 4 = 10
...
This is an additive factorial, as it were. It can be expressed as:
N(s) = Σ (from i = 1 to s) of i
Reading through the replies, ah yes, Gauss solved this centuries ago.
Tim
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franklyn
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More trickery
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Pyro
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Ok Magpie. My maths exam had a similar question.
U(n)=U(n-1)+1
or:
U(n)=U1+(n-1)
and to get the total amount:
Sn=(U1+Un)/2
all above information is intellectual property of Pyro.  |
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Eddygp
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Given two equal numbers:
a=b; multiply by "a" each side
a^2=ab; subtract b^2 from each side
a^2-b^2=ab-b^2; which is the same as:
(a+b)(a-b)=b(a-b); divide by (a-b)
a+b=b; now, as a=b, substitute:
2b=b; divide by b
2=1
there may be bugs in gfind
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gsd
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Quote: Originally posted by Eddygp  | Given two equal numbers:
a=b; multiply by "a" each side
a^2=ab; subtract b^2 from each side
a^2-b^2=ab-b^2; which is the same as:
(a+b)(a-b)=b(a-b); divide by (a-b)
a+b=b; now, as a=b, substitute:
2b=b; divide by b
2=1
|
This is too matured a forum for this type of 4th grader cleverness.
If a=b then your can't divide by a-b
gsd
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Eddygp
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| Quote: Originally posted by gsd | | Quote: Originally posted by Eddygp | Given two equal numbers:
a=b; multiply by "a" each side
a^2=ab; subtract b^2 from each side
a^2-b^2=ab-b^2; which is the same as:
(a+b)(a-b)=b(a-b); divide by (a-b)
a+b=b; now, as a=b, substitute:
2b=b; divide by b
2=1
|
This is too matured a forum for this type of 4th grader cleverness.
If a=b then your can't divide by a-b
gsd |
I just found it funny.
there may be bugs in gfind
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franklyn
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AJKOER
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Area of Triangle = 1/2 ab*sin(angle between sides a & b)
Now, we are given that a and b are the same in the new triangle by construction, but for a triangle with apparently less area.
Or, substituting in our formula for area:
1/2 ab*sin(angle between sides a & b in 2nd triangle) < 1/2 ab*sin(angle between sides a & b in 1st triangle)
Cancelling like terms and taking the inverse of the sine function implies that:
Angle between sides a & b in 2nd triangle < Angle between sides a & b in 1st triangle
as was claimed to be the explanation.
[EDIT] This trick is not much of a trick if one reflects each triangle to form parallelograms. If one sits on a parallelogram, the length of the sides
remains constant, but the area changes (obviously).
---------------------------------------
New Question: Your big fat in-law sits on a perfectly semi-circular cushion whose base is wood (think of it as a very flexible half sphere connected
to a fixed base).
How has the volume of the cushion changed?
What if his ultra-thin wife sits on it?
State your assumptions and prove your results.
[Edited on 30-3-2013 by AJKOER]
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Eddygp
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www.coopsjokes.com/toons22/f-64is65.htm
WHAT?
there may be bugs in gfind
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Pyro
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that was discussed earlier. also in the ''need a laugh'' thread
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Eddygp
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yes, that is where I saw it... kinda looks better in this thread though.
there may be bugs in gfind
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Eddygp
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http://24.media.tumblr.com/9bff5678b2f5ffcd18fba31b21f005c1/...
there may be bugs in gfind
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franklyn
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Two plus eleven = one plus twelve , both sides contain the same thirteen letters , e, e, e, l, l, n, o, p, s, t, u, v, w
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