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math puzzle

Magpie - 19-5-2007 at 14:26

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.

pantone159 - 19-5-2007 at 14:49

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]

Magpie - 19-5-2007 at 15:08

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.

pantone159 - 19-5-2007 at 17:22

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.

Magpie - 19-5-2007 at 19:11

Quote:

Re the formula you looked up, do you mean for 1+2+3+...+s ?

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.

JohnWW - 19-5-2007 at 21:56

Chapter 2 of Perry's Chemical Engineers' Handbook also has the formulae.

woelen - 20-5-2007 at 06:34

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.

Basic Algebra

MadHatter - 20-5-2007 at 12:11

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

Magpie - 20-5-2007 at 20:06

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]

math puzzle soln.jpg - 38kB

franklyn - 23-5-2007 at 15:21

Here's one that's not obvious

Hint : don't assume anything !

illusion_how.gif - 19kB

chemoleo - 23-5-2007 at 16:54

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!

Magpie - 23-5-2007 at 19:43

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]

12AX7 - 24-5-2007 at 16:03

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

More trickery

franklyn - 28-3-2013 at 08:55

Quote: Originally posted by Eddygp in Whimsey

I got dazzled: http://www.coopsjokes.com/toons22/f-64is65.htm

Pyro - 28-3-2013 at 10:12

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

Eddygp - 30-3-2013 at 02:23

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

gsd - 30-3-2013 at 03:44

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

Eddygp - 30-3-2013 at 04:20

Quote: Originally posted by gsd

Quote: Originally posted by Eddygp

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.

franklyn - 30-3-2013 at 08:37

AJKOER - 30-3-2013 at 09:54

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]

Eddygp - 30-3-2013 at 10:58

www.coopsjokes.com/toons22/f-64is65.htm

WHAT?

Pyro - 30-3-2013 at 13:21

that was discussed earlier. also in the ''need a laugh'' thread

Eddygp - 30-3-2013 at 13:26

yes, that is where I saw it... kinda looks better in this thread though.

Eddygp - 5-4-2013 at 13:21

http://24.media.tumblr.com/9bff5678b2f5ffcd18fba31b21f005c1/...

franklyn - 30-8-2014 at 00:40

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

Eddygp - 30-8-2014 at 06:14

Quote: Originally posted by franklyn

Here's one that's not obvious

Hint : don't assume anything !

SPOILER:
_{Both triangles are different}