Math Is Fun Forum

If you all are interested in logical puzzles, here's one:-

A Captain has three smart soldiers under him. The three soldiers are informed that their intelligence is being tested and that the captain has three blue and two red caps. They are made to stand in ascending order of their heights. They are blindfolded and the captain puts a cap, chosen at random, on the heads of the three soldiers. The blindfold is then removed. The captain asks the third soldier whats the color of the cap on his head. The third soldier sees the color of the cap on the heads of the first two soldiers, but he says he is unable to tell the color of the cap on his head.
The Captain asks the same question to the second solider. The second solider takes into consideration the third soldier's reply and the color of the cap on the first soldier, which he is able to see. But, he too is unable to tell the color of the cap on his head.
The Captain then asks the same question to the first soldier. The first soldier has only the replies of the third and second soldiers to draw any conclusion. But he tells the color of the cap on his head. He gives the right answer.

What's the color of the cap on the first soldier's head and how did he answer correctly?

Normally, in such paradoxes, this is where the mistake lies.
Since a² = b²,
it cannot be inferred that a=b.
Taking square root on both sides of an equation should always have the
'±' sign, after the square root is worked out.

That is, it is to be proved that
(abcd..........)^n = a^n+b^n+c^n+d^n..................
Here, both a,b,c,d...etc and n go on and on.
It is easy to say there would always exist a n digit number whose sum of digits raised to the power n is equal to the number. To prove that may be quite difficult. As n becomes higher, the combinations available increase, thereby increasing the possibility of such a number existing.

Good, you liked the website, Tara Witney!
You are wlecome to join the forum!

Great work, Mathsy!
I don't know how you solved the problem, but this is how I would do it.

Let 'l' denote the light that strikes the surface of water at depth 'h' then
dl/dh = -kl
dl/l = -kdh
Integrating,
log l = -kh + c
Let log l = lo when h=0, so that
log (l/lo) = -kh
l/lo = (100-40)/100 = 3/5 when h=10
therefore, log 3/5 = -10k --------------------------(I)
We have to find h when
l/lo = 1/300,000
log (1/300,000) = -hk -----------------------------(II)
Dividing II by I,
-log 300,000/ log 0.6 = h/10,
h=247 metres approximately.

Problem # n+1
We know water absorbs light by the fact that it is dark in the ocean depths. If 10 metres of deep water absorbs 40% of the light which strikes the surface, at what depth would light striking the surface be reduced to 1/300,000 th? Assume that the rate of absorption is proportional to the light.

Can you post the proof? I got as far as Mathsy did, that two of the numbers should be odd and one even. Maybe, the proof has got something to do with 2 being the only even prime

Mathsisfun has proved that he is a Computer Scientist

I had read somewhere that animals (the most intelligent ones) cannot count beyond 7 or 8! They are good at counting up to 4, thereafter, there is a progressive decline in their ability to count, and beyond 8, they are unable to count!

With the help of the formula
c² = a² + b² - 2ab Cosθ,
you can find the value of c.
Now, use the formula
b² = a² + c² - 2ac Cosθ.
You know the value of a,b, and c.
You would get
Cosθ = x (some value)
The angle opposite to side 'b' would then be
θ = Cos-¹x.
Is that clear?

Zmurf,
This formula is easier remembered as

a² = b² + c² - 2bc cosθ
On simplification of the equation given by you, this is what is obtained.
a,b, and c are the three sides of a  triangle.
If the triangle is rightangled and θ is 90 degrees or pi/2 radians,
Cos θ = 0,
which gives us
a² = b² + c²,
the Pythogoras Theorem.

You are right, Rora! Mathsisfun is remarkable