[Arrange numbers to the equations, so if we turn the equations upside down, they are still right]

_ + _ = _

_ + _ = _

_ + _ = _

_ + _ = _

 

Fill the blanks with numbers above, so:

• The 4 equations are right.
• All numbers must be used.
• If we turn the equations upside down, they are still right.
• You can rotate any numbers too, example 1091  ->  1601.

[Arrange numbers and operators to the magic triangle]

When creating this puzzle (using computer), I started with 3 first constraints, then I got many solutions, so I have to add 1 more constraint. But I have to keep the symmetry of the puzzle. First I tried to sum the three corners, but I still got some solutions. Then  I tried to product the three corners, which yields 1 unique solution.

When creating this puzzle (using computer), I started with 3 first constraints, then I got many solutions, so I have to add 1 more constraint. But I have to keep the symmetry of the puzzle. First I tried to sum the three corners, but I still got some solutions. Then  I tried to product the three corners, which yields 1 unique solution.

[Replace letters with numbers 1 to 9, so the 3 operations are equals]

Replace letters with numbers 1 to 9, so the 3 operations below are equals. Each letter represents one unique number.
  ab / c = de * f = gh - i

Note : 

• If a = 1, and b = 2, then ab = 12  NOT multiplication
• It is more fun if you solve this without computer, trust me this is easy.

[Find the number (my version)]

based on this @bonanova puzzle, I create another similar puzzle

You are given the following ten statements and are asked to determine a particular number.

1. At least one of statements 7 and 8 is true.
2. This either is the first true or the first false statement.
3. The number is a prime number.
4. The first true statement multiplied by the last false statement divides the number.
5. The number of divisors of the number is greater than the sum of the numbers of the true statements.
6. The number has exactly 4 prime divisors.
7. The number is bigger than 1000.
8. The numbers of true statements do not equal the numbers of false statement
9. One of the divisors is a cube number bigger than 1.
10. There are 3 consecutive False statements and 3 consecutive True statements.

[Fun with digital sums]

yes there are still 2 more solution.

Spoiler

 

1458 = 18 * 81

and 1729 = 19 * 91

No more solution, because for numbers bigger than that (addition of digits) * (reverse of addition of digits)  will too small for the numbers.

 

 

[Arrange numbers and operators to the magic triangle]

6 hours ago, bonanova said:

Clarification?

Each variable {A B C D E F G H I} can be used only once.
I now assume, but it's not clear, that each operator {+ - x / ^ 10A+B]} can be used only once, making (9!)(6!) cases.

I started allowing any operator to be used in each white triangle.

Yes you are right, each operator  only can be used once. making ((9!)(6!))/3 cases. (Because of the symmetry)

[Arrange numbers and operators to the magic triangle]

 

Note : A # B = A*10 + B

Arrange the numbers 1 to 9 to green triangles, and arrange operator (+,-,x,/,^, and #) to the white triangles, so the math operations below are equals.

• (((A op1 B) op2 D) op4 F) = constant
• (((F op4 G) op5 H) op6 I) = constant
• (((I op6 E) op3 C) op1 A) = constant
• A x F x I = constant

[Reverse the order of 8 poisonous substances]

10 hours ago, rocdocmac said:

Does the order of robot function remain constant, i.e. Robot 1, Robot 2, Robot 3, Robot 1, Robot 2, Robot 3, Robot 1, ....?

No, you can choose the robot in each step something like 2213232.....

[Interesting 3x3 table, to form a numbers of unique prime numbers]

On 11/6/2016 at 2:39 PM, tojo928 said:

Fun extra question to add in to challenge on this one. What is the lowest number of unique primes that could be used in this and how many unique ways (including rotation and reflection) can this be made?

Spoiler

 

3 unique primes, there are some solution for 3 unique primes, one of them is 

1 3 1     

3 5 3

1 3 1

 

 

[Reverse the order of 8 poisonous substances]

8 very poisonous substances named s1 to s8 are kept in a safety room.
The substances are kept in ascending order (s1,s2,s3,s4,s5,s6,s7 and s8).
In the room there are 3 robots.

First robot can "rotate left" the order of the substances.
If the order is (a,b,c,d,e,f,g,h) the robot will make it (b,c,d,e,f,g,h,a)

Second robot can split the substances into 2 part then reverse the order of each part.
If the order is (a,b,c,d,e,f,g,h) the robot will make it (d,c,b,a,h,g,f,e)
But the robot is a bit broken, so the resulting order is a bit wrong.
The resulting order will become (d,c,a,b,h,g,f,e).

Third robot can split the substances into 4 part then reverse the order of each part
If the order is (a,b,c,d,e,f,g,h) the robot will make it (b,a,d,c,f,e,h,g)

Questions

If the 2nd robot is not broken, how many minimum step needed by using the robots to reverse the order into descending order? Show the steps!

After the 2nd robot is broken, how many minimum step needed by using the robots to reverse the order into descending order? Show the steps!

[Interesting 3x3 table, to form a numbers of unique prime numbers]

It is possible, I have it. I will show it next week if nobody find it until 10 Nov 2016.

6 hours ago, tojo928 said:

  Hide contents

Not possible to get 9 unique values

 

 

[Interesting 3x3 table, to form a numbers of unique prime numbers]

 

 151  131 359 131  151
    \   ^   ^   ^ /
  131 < 1   3   1 > 131
  353 < 3   5   3 > 353
  191 < 1   9   1 > 191
    /   V   V   V \  
  151  131 953 131  151

This 3x3 table have an interesting properties. 
Every direction (up,down, Right, Left, 45°,135°,225°,315°) of 3 cell form prime numbers.
There are 6 unique prime numbers from this table, they are 131, 151, 191, 353, 359, 953.

Create a more interesting 3x3 table with the same rule, which there are 9 unique prime numbers from the table.

[Arrange the numbers 1 to 19 in the circles]

     O---o---O
    / \     / \
   o   o   o   o
  /     \ /     \
 O---o---O---o---O
  \     / \     /
   o   o   o   o
    \ /     \ /
     O---o---O

Arrange the numbers 1 to 19 in the circles, so all the rows of 3 numbers between O (big o) sums to 23

Example :

     a---b---c
    / \     / \
   d   e   f   g
  /     \ /     \
 h---i---j---k---l
  \     / \     /
   m   n   o   p
    \ /     \ /
     q---r---s

a+b+c = a+e+j = c+f+j = 23, and so on....

Note :

• There are 2 solutions, if we rule out reflections and rotations.

Bonus Puzzle : How if all the rows of 3 numbers between O (big o) sums to 22,24,25,26,27,28,29,31,32,33,34,35,36, 37, and 38. I have checked all the solutions by computer, and all those sums have solutions.

[Interesting pattern on 5x5 table which has unique properties]

[0,1,6,4,3]
[4,5,6,0,9]
[9,9,0,1,1]
[1,0,4,5,6]
[7,6,4,9,0]

This 5x5 table has unique properties.
Each number in a cell means :

The cell = last digit of (sum of its neighbour (including diagonals))

or

The cell = The remainder of the sum of its neighbour divided by 10

Here is another example

 [0,1,6,4,8]   [1,2,1,3,2]   [1,0,9,0,1]
 [4,5,6,0,4]   [4,5,1,0,9]   [6,5,9,5,6]
 [4,4,0,6,6]   [3,3,0,7,7]   [5,5,0,5,5]
 [1,0,4,5,1]   [1,0,9,5,6]   [9,0,1,0,9]
 [2,1,4,4,0]   [8,7,9,8,9]   [4,5,1,5,4]

What surprised me is, every table like this will follow this :

• The middle cell is always 0.
• Any other cell (i,j), (6−i,6−j) adds upto 0 modulo 5 , that means, (i,j) + (6−i,6−j) is completely divisible by 5. 

I have checked this with my computer.

Why this happens ?

[ENLIST + SILENT + LISTEN = ANAGRAM]

Solve this alphametic

ENLIST + SILENT + LISTEN = ANAGRAM

Leading zero is ok.
There is only 1 solution.

[alphametic with modulo]

Solve this alphametic (replacing letters with digits)

ABCDEFG mod (A×B×C×D×E×F×G)=0

example :

ABCDE mod (A×B×C×D×E)=0
13248 mod (1×3×2×4×8)=0

Note :

• Do not replace any letter with 0.
• Each letter represent different digit.
• There is only one solution.

[Star Puzzle : Determine which circles are True and which circles are False]

 

A circle with number x means :
“This circle is connected (straight line) to x circles with true statement”
This is valid only if the statement in the circle is True.

If a circle is False, then it could be interpreted like this:
“This circle is NOT connected (straight line) to x circles with true statement”
Thus, the number of True statements connected to a False circle should be different than the number in the False circle itself.

Let’s say a true circle is a circle with a true statement.
So if the number is 3 means the circle is connected by 3 true circles.
Some circles are true and some are not. 
Each circle is connected to 5 other circles, except the middle circle.
The middle circle is connected to 10 other circles.

Create another star with Boolean (T/F) input,
to show which circles are true, and which circles are false.

Example

Note

A is connected to (B,C,D,E,F)
B is connected to (A,C,F,G,J)
C is connected to (A,B,D,F,G)
D is connected to (A,C,F,H,E)
E is connected to (A,D,F,H,K)
F is connected to (A,B,C,D,E,G,H,I,J,K)
G is connected to (B,C,F,I,J)
H is connected to (E,D,F,I,K)
I is connected to (J,G,F,H,K)
J is connected to (B,G,F,I,K)
K is connected to (J,I,F,H,E)

[Guessing right number from 1 to 27, with 3 Yes/No/Maybe questions.]

 

Spoiler

Here is my version :

I  will create 3 pairs of list

First pair : 
1A : 2,3,5,6,8,9,11,12,14,15,17,18,20,21,23,24,26,27
1B : 2,5,8,11,14,17,20,23,26

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in first paper.
(because the number appears in both list)
If the answer is No then, write number 0 in first paper.
(because the number do not appears in both list)
If the answer is I do not know then, write number 1 in first paper.
(because the number only appears in one list)

then

2nd pair :
2A : 4,5,6,7,8,9,13,14,15,16,17,18,22,23,24,25,26,27
2B : 4,5,6,13,14,15,23,24,25

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in the 2nd paper.
If the answer is No then, write number 0 in the 2nd paper.
If the answer is I do not know then, write number 1 in the 2nd paper.

then

3rd pair :
3A : 10 to 27
3B : 10 to 18 

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in the 3rd paper.
If the answer is No then, write number 0 in the 3rd paper.
If the answer is I do not know then, write number 1 in the 3rd paper.

then

put the number in the order 3rd 2nd 1st as base 3 number, than convert it into base 10 number,do not forget to add 1. The result is Mr. Smith number.

 

 

[How many spin-able numbers are there from 0 to 99999 ?]

0,1,2,5,8,11,69,96 are spin-able numbers.

We call a number is a spin-able number if :

When it is typed on a calculator, then spin the calculator 180 degree, the number is still the same number with the original number.

How many spin-able numbers are there from 0 to 99999 ?

This is a no-computer puzzle, only right answers with explanation will be accepted.

[Guessing right number from 1 to 27, with 3 Yes/No/Maybe questions.]

Your best friend (Bob) comes to you with this problem.

The Problem

His teacher (Mr. Sam) will ask Bob to work out a number in a child (Joan) head from 1 and 27.
 Bob has to find the correct number, by asking Joan three Yes/No questions about the number.
 Joan can respond with 3 different answers: "Yes", "No", or "I do not know".

Joan is a smart boy, but he doesn't know about another base number except 10. So by asking that kind of question he will answer "I do not know"

Example:
If the number is 12

Is it an even number? Yes.
 Is it divisible by 9? No.
 If I take a random number between 10 to 15, will I get the right number? I do not know.

------------------

Another Example:
If the number is 9

Is it an even number? No.
 Is it divisible by 9? Yes.
 If I take a random number between 10 to 15, will I get the right number? No.

------------------
 

Find the strategy, so every number from 1 to 27 can be guessed correctly.
 

[5x5 statement table.]

@ bonanova

If you mean same puzzle with different numbers, you can create  It is easily, but maybe the solution is not unique.

if you mean different puzzle with T/F statement, I do not have, but I wiil create it, with another shape, maybe hexagon, triangle, or  etc.

[5x5 table with unique properties]

1 3 2 3 1
3 3 2 3 3
2 2 3 3 2
3 3 3 4 4
1 3 2 4 1

This 5x5 table has unique properties.
Each number (x) in a cell means : 

The cell is surrounded (including diagonal neighbours) by x unique numbers in the squares surrounding it.

 

Example :
(2,2) cell (red) is 3, so it is surrounded by 3 unique numbers (1,2 and 3)
(4,4) cell (blue) is 4, so it is surrounded by 4 unique numbers (1,2, 3 and 4)

The Question

3 . . . .
. . . . 2
. . . . .
2 . . . .
. . 3 . .

Fill in the blanks so the table has the same unique properties.

[5x5 statement table.]

@ Phill, to make the question clear, I will explain the example

1 3    ->  T F
0 1        F T

 

(1,1) is true because the cell, is surrounded by 1 true statement (2,2)

(1,2) is false because the cell is not surrounded by 3 true statement, it only surrounded by 2 true statement (1,1) & (2,2)
(2,1) is false because the cell is not surrounded by 0 true statement, it surrounded by 2 true statement (1,1) & (2,2)
(2,2) is true because the cell, is surrounded by 1 true statement (1,1)

[5x5 statement table.]

@bonanova : yes diagonals count.

[5x5 statement table.]

Statement table.

3 2 1 4 1
2 3 3 3 3
3 4 6 4 4
2 3 4 4 3
1 1 2 4 3

Above 5x5 table contains numbers.
A number at each cell represent a statement.

x : This cell is surrounded by x True statements.

So if the number is 3 means the cell is surrounded by 3 True statement
We can say all statemets are false, but this is not what I want.
Some statements are true and some are not.

Create another 5x5 table with boolean (T/F) input,
to show which statements are true, and which statements are false.

example for 2x2 table.

1 3    ->  T F
0 1        F T

I have checked there is only 1 solution.
Find the solution !