Riddles, brain puzzles and mathematical problems

[BAndrew]

@Oscar House

That's it! You nailed it

[BAndrew]

Here's more:

• Prove that all primes greater than 2 can be written as the difference of two squares.
  
  
• Decode this:
  31 51 02 51 81 90 91 02 91 10
  81 50 10 91 11 50 40 02 51 20
  50 51 41 02 80 50 21 51 51 11
  51 12 02 60 51 81 91 90 42 02
  50 50 41 80 10 81 40 50 41 50
  40 30 81 90 31 90 41 10 21 91
  
  
• Your sister and your best friend have been acting rather suspiciously recently. One evening you notice your sister secretly pass a piece of paper into your friend's pocket. You wait for the right opportunity and remove the note to discover the following:
  
  
  R1U0R0L4L6D5L7D1 L4U6R0R2U4 U6L0D8 R1D1R2.
  R1U0L7L4D1U8R2R1 L2L3L8D4D5L2D3!
  
  Confused, you record the secret message and return the piece of paper. The next day, while your sister is out, you look around the house and discover a scrap of paper in one of the waste paper bins. It has the following similar message written on it, but it seems to contain a clue:
  
  
  R2D1R1L8 = TEST
  
  Can you decode the original message and work out what your sister and best friend are up to?
  
  
• Prove that n must be odd for (1+1/2)(1+1/3)(1+1/4)...(1+1/n) to be an integer.
  
  
• Find the next number:
  2,3,5,7,11,13,?
  

[BAndrew]

BONUS:

A strange cult, with a belief that strawberries contain the Elixir of life, have succesfully isolated the gene responsible for producing the flavour in strawberry plants. However, it seems that they have been able to combine this material with bovine DNA to produce a highly infectious virus that affects cattle. Although the viruscauses no harm to the cow, it will cause the milk that it produces to be tainted with a strawberry flavour. Once infected there is no cure and all future generations will produce stawberry flavoured milk. We were able to intercept an encoded message, sent to all members of the cult, regarding a final meeting to discuss the release of the virus. We know that the meeting has been arranged to take place in Athens,
Greece. It is imperative that you decode this cipher so that we can uncover the identity of the mysterious leader of this cult and arrange his arrest.

LFI TIVZG OVZWVI, WLMZOW NXILMZOW, RMERGVH ZOO NVNYVIH GL GSV
URMZO NVVGRMT RM ZGSVMH, TIVVXV.

[BAndrew]

1)Fun fact:

Consider the following function:

f(n) = sqrt(9/121*100^n +(112-44n)/121)

Find for fun (with the help of a computer) f(95). Notice something strange?

Calculate 10000 or so digits to see something cool.


2)Math riddle:
Find the positive integers x,y,z so that
x^2 + y^2 + z^2 = N^3
x^3 + y^3+ z^3 = M^2

(at least 1 solution required + how you found it)
Are there infinite solutions?

3)IMPOSSIBLE(?) PROBLEM

Find the smallest number with the following property:

When devided with AB, ABA, ABAB the remainder is A, AB, ABA respectively (Letters A and B symbolize the digits of the number). You have AB days to solve the problem, otherwise ABA million humans will be killed and you will have to pay AB^10 $.

[Wapez]

Fun fact:

1 + 2 + 3 + 4 +5 ... = 1/12

The sum of all positive numbers up to infinity, is equal to 1/12.

[BAndrew]

Fun fact:

1 + 2 + 3 + 4 +5 ... = -1/12

The sum of all positive numbers up to infinity, is equal to -1/12.

Not exactly...

The sum of all the positive numbers does not converge. It is divergent series. However, in specialized mathematical contexts, values can be usefully assigned to certain series whose sequence of partial sums diverges. A summability method or summation method is a partial function from the set of sequences of partial sums of series to values.

[BAndrew]

Easy:

There are 12 coins. One of them is false; it weights differently. It is not known, if the false coin is heavier or lighter than the right coins. How to find the false coin by three weighs on a simple scale?

[Froge]

Prove that n must be odd for (1+1/2)(1+1/3)(1+1/4)...(1+1/n) to be an integer.

Denote X as the set of natural numbers greater than or equal to 2.

P(n):

(1+1/2)(1+1/3)...(1+1/n) ∈ Z <-> n = 2k+1, k ∈ Z, k >= 1 for n ∈ X. Proof by induction.

P(1): (1 + 1/2) = 3/2, not an integer, but: (1 + 1/2)(1 + 1/3)
= 1 + 1/2 + 1/3 + 1/6
= 2 is an integer.

Suppose P(k) is true for k ∈ X. If k is odd:

(1+1/2)....(1+1/k) = A ∈ Z.

Consider now k+1, which is even:
(1+1/2)....(1+1/k)(1+1/(k+1)) = A(1 + 1/(k+1))
Given 1 + 1/(k+1) cannot be an integer for k ∈ X, A(1+1/(k+1)) is thus not an integer.

Consider now k+2, which is odd:
(1+1/2)....(1+1/k)(1+1/(k+1))(1+1/(k+2)) = A(1 + 1/(k+1))(1+1/(k+2))
=A(1 + 1/(k+1) + 1/(k+2) + 1/(k+1)(k+2))
=A(1 + (k+2 + k+1 + 1)/((k+1)(k+2)))
=A(1 + (2k + 4)/((k+1)(k+2)))
=A(1 + 2(k+2)/((k+1)(k+2)))
=A(1+ 2/(k+1))
=A((k+3)(k+1))

....I'm stuck, but I suspect a similar procedure follows for if k is even, and then by principle of mathematical induction, P(n) is true.

[BAndrew]

Hint: Don't use mathematical induction. You can probably solve it, but it's way harder (+ there could be some flaws in your logic if you aren't careful because you aren't trying to prove something like (prove 1+2+3+..+n = n*(n+1)/2), but you want to prove that A is integer when n is odd). Try writing 1 + 1/n as (n+1)/n
The answer is 2 lines long (depends on how you write of course). With induction (as you saw) it already became complicated.

[Romulator]

Your sister and your best friend have been acting rather suspiciously recently. One evening you notice your sister secretly pass a piece of paper into your friend's pocket. You wait for the right opportunity and remove the note to discover the following:

R1U0R0L4L6D5L7D1 L4U6R0R2U4 U6L0D8 R1D1R2.

R1U0L7L4D1U8R2R1 L2L3L8D4D5L2D3!

Confused, you record the secret message and return the piece of paper. The next day, while your sister is out, you look around the house and discover a scrap of paper in one of the waste paper bins. It has the following similar message written on it, but it seems to contain a clue:

R2D1R1L8 = TEST

Can you decode the original message and work out what your sister and best friend are up to?

I won't lie, first thought she was becoming a lesbian

Thinking about R, since there are two, we have R2 and R1. Comparing them in position relative to TEST, 2 places after R in the alphabet is T, and 1 after is S. Similar circumstances for the others, with one letter after D = E, and 8 letters after L = T - likely to put us off.

Using this:

Spoiler below!

Code: (Select All)

//Reference

ABCDEFGHIJKLMNOPQRSTUVWXYZ

//Message
R1 = S
U0 = U
R0 = R
L4 = P
L6 = R
D5 = I
L7 = S
D1 = E

L4 = P
U6 = A (Go back to A, as Z is 5 letters after U)
R0 = R
R2 = T
U4 = Y

U6 = A
L0 = L
D8 =  L

R1 = S
D1 = E
R2. = T.


R1 = S
U0 = U
L7 = S
L4 = P
D1 = E
U8 = C (Like U6, go back to A. Becomes C)
R2 = T
R1 = S

L2 = N
L3 = O
L8 = T
D4 = H
D5 = I
L2 = N
D3! = G!

Therefore, the secret message is:

SURPRISE PARTY ALL SET.
SUSPECTS NOTHING!