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2, 0, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, ...

What are the next five numbers, and why?

(No Solution Yet, 0 Comments) Submitted on 2017-09-20 by Charlie    

Before trying the problem "note your opinion as to whether the observed pattern is known to continue, known not to continue, or not known at all."

After row 1, row n is obtained by inserting n between every pair of consecutive numbers that sum to n.

row 1:                  1 1
row 2:                 1 2 1
row 3:               1 3 2 3 1
row 4:            1 4 3  2  3 4 1
row 5:        1 5 4  3 5 2 5 3 4 5 1
row 6:   1  6  5 4  3 5  2  5  3  4 5  6 1
row 7: 1 7 6 5 4 7 3 5 7 2 7 5 3 7 4 5 6 7 1
The number of numbers in each row are 2, 3, 5, 7, 11, 13, 19...

Will they always be prime?

(No Solution Yet, 3 Comments) Submitted on 2017-09-19 by Jer    

Consider two points on parabola y=x2, (-a,a2) and (b,b2), where a and b are distinct real numbers.

If these two points are connected by a straight line, where does that line intersect the y-axis?

Inspired by an interactive sculpture at the Museum of Mathematics, NYC.
(No Solution Yet, 2 Comments) Submitted on 2017-09-18 by Charlie    

Let ABC be a right-angled triangle with hypotenuse c = AB.
Let Wa and Wb be the lengths of the angle bisectors from A & B, respectively.

Prove that Wa + Wb ≤ 2c*sqrt( 2−√2).

(No Solution Yet, 0 Comments) Submitted on 2017-09-17 by Ady TZIDON    

Before trying the problems "note your opinion as to whether the observed pattern is known to continue, known not to continue, or not known at all."

Is 7013*2n+1 always composite?

Is 78557*2n+1 always composite?

(No Solution Yet, 4 Comments) Submitted on 2017-09-16 by Jer    

After successfully solving an alphametics w1+w2=w3,where those three words spelled out the surnames of three famous Scottish writers WG sent the solution (216035+21754=237789) to his colleague, requesting him to reconstruct the original names and to mail his answer ASAP.
24 hours later a short message arrived: "A nice brush!"

Your comments are welcome...

(No Solution Yet, 2 Comments) Submitted on 2017-09-15 by Ady TZIDON    

Difficulty: 2 of 5 Cryptaquote I (in Word Problems) Rating: 5.00
"A B and A C are very rare".

Either decode the famous quotation or invent a meaningful one of your own.

A is an adjective, B and C are nouns.
(No Solution Yet, 6 Comments) Submitted on 2017-09-14 by Ady TZIDON    

In his paper The Strong Law of Small Numbers Richard Guy states "There aren't enough small numbers to meet the demands made of them."

It's a great list of 35 examples where the pattern noted early on may or may not continue. Unfortunately, if you read it, you will give away a series of around 10 puzzles I plan to create from it.

Before trying the problem "note your opinion as to whether the observed pattern is known to continue, known not to continue, or not known at all."

Place n points around a circle so that no three of the C(n,2) chords joining them are concurrent. Count the number of regions into which the chords partition the circle.

n=0, 1 region
n=1, 2 regions (a single chord)
n=2, 4 regions (the chords form a triangle)
n=3, 8 regions
n=4, 16 regions

A pattern has emerged. Does it continue?

(No Solution Yet, 8 Comments) Submitted on 2017-09-13 by Jer    

Take a sequence of non-negative integers, i.e. 0,1,2,3, . etc
and erase all square numbers.
You will be left with a sequence like : 2,3,5,6,7,8,10 ...etc.

Find an explicit formula for an.

(No Solution Yet, 1 Comments) Submitted on 2017-09-12 by Ady TZIDON    

A standard alphametic:

x    I
(No Solution Yet, 4 Comments) Submitted on 2017-09-11 by Jer    

Difficulty: 3 of 5 Lots of lods (in Sequences) Rating: 4.67
Let's denote the largest odd divisor of a positive integer n by lod(n).
Thus lod(1)=1; lod(72)=9; lod(2k+1)=2k+1; lod(2^k)=1.

Prove the following statement:
Sum of all values of lod(k), (k>1) from k=n+1 to k=2n, inclusive, equals n2.

Example: take n=7: lod(8,9,10,11,12,13,14)= (1,9,5,11,3,13,7), sum of the values within the last pair of brackets is 49, indeed.

Source: Shown to me as a trick i.e. "the wizard" guesses the result a priori.

(No Solution Yet, 7 Comments) Submitted on 2017-09-10 by Ady TZIDON    

Let ABC be an acute triangle with altitudes AD and BE. The
intersection of AD and BE is H. Points F and G make CADF
and CBEG into parallelograms. M is the midpoint of FG.
Ray MC intersects the circumcircle of ΔABC again at point N.

Prove that ANBH is a parallelogram.
(Solution Posted, 4 Comments) Submitted on 2017-09-09 by Bractals    

Six persons lived in a castle wherein each occupant had his own apartment. One morning, two of the occupants were dead:

A died because he sat on a poisoned needle that was hidden in his bed. In the last moment of his life, he scratched into the bedsheet with the needle that he took out of his backside: M = C
It was understood to mean:

A: "My murderer is C."

B was suffocated by a tampered Baroque bed canopy that crashed down to him. In the last seconds of his life, he carved a message on the wooden bedstead with a cork-screw in the one hand hanging out:

B: "My murderer is F."

Collection and evaluation of evidence by the police proved that each victim had been murdered by one of the other occupants. C, D, E, and F refused to answer police questions. There was hope that they would speak at the trial. But since the court room was bombed near the end of the proceedings and no one survived inside, the outcome of the court case remained unknown until now.
Seventy years later, two half burnt pieces of the court record were found in a huge forgotten box full of stones and ashes. The first piece of paper contains the statements of C, D, E, and F:

C: "F is a murderer or D is lying as always."

D: "A's or B's statement is erroneously false, or C's statement is true."

E: "If what I say is true, then the murderer of A and the murderer of B are not the same."

F: "If C murdered B or E is telling the truth, then I killed A."

The second piece of paper contains a snippet of the court decision:

"...the A case and the B case are closed and confirm the rule: No murderer without a lie..."

Can you prove that there is a unique solution for one of the victims and that there are two solutions for the other one?

(The word "or" was used to mean "at least one, and maybe both".

(No Solution Yet, 0 Comments) Submitted on 2017-09-08 by ollie    

Show that (1!*2!*3!*4!* ... 100!)/50! is a square number.
(No Solution Yet, 8 Comments) Submitted on 2017-09-07 by Ady TZIDON    

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