10/31/2015
What do you get when you take the middle out of a hotdog on the last day of October?
A Hollow Weiner!
David
Saturday, October 31, 2015
Every Number has a Story to Tell: 43,274,246
10/31/2015
Every Number has a Story to Tell:
Today’s Story is told by 43,274,246.
43,274,246 is a composite, even and square
free number.
The product of its digits is 32256, while
the sum of its digits is 32.
The prime factors of 43,274,246 are 2, and 21637123. It is a semiprime because it is the product
of two primes. The sum of its prime
factors is 21637125.
The divisors of 43,274,246 are 1, 2,
21637123, and 43,274,246, whose sum is σ = 64911372. Its totient is φ =
21637122.
It is a congruent number.
It is not an unprimeable number, because it
can be changed into a prime (43274243) by changing a digit.
It is a polite number, since it can be
written as a sum of consecutive naturals, namely, 10818560 + 10828561 +
10818562 + 10818563.
It is an arithmetic number, because the
mean of its divisors is an integer number (16227843).
Almost surely, 2^{43,274,246} is an
apocalyptic number.
43,274,246 is a deficient number, since it
is larger than the sum of its proper divisors (21637126).
43,274,246 is a wasteful number, since it
uses less digits than its factorization.
43,274,246 is a number that does not occur
in the OEIS data base.
43,274,246 is an evil number, because the
sum of its binary digits is even.
It is the short leg of a Pythagorean
Triangle: 43,274,246^{2} + 771551240^{2} = 771551242^{2}.
The string “43274246” occurs at position
21104172 in the decimal expansion of pi (counting from the first digit after
the decimal point, the “3.” is not counted).
It occurs only once in the first 200 million digits of pi.
The Cosine value of the number 43,274,246
is 0.53861073324193.
The Hyperbolic Cosine value of the number 43,274,246
is INF.
The Binary value of the number 43,274,246
is 10100101000101000000000110.
The Hexadecimal value of the number 43,274,246
is 2945006.
The Octal value of the number 43,274,246 is
245050006.
The Degrees to Radian value of the number 43,274,246
is 755278.07401799.
The Exponent value of the number 43,274,246
is INF.
The Logarithm value of the number 43,274,246
is 17.58306823542.
The Radian to Degrees value of the number 43,274,246
is 2479431657.4109.
The Sine value of the number 43,274,246 is
0.84255473296196.
The Hyperbolic Sine value of the number 43,274,246
is INF.
The Tangent value of the number 43,274,246
is 1.5643110709855.
The Hyperbolic Tangent value of the number
43,274,246 is 1.
But the thing that very few people know
about this number:
43,274,246 is a composite number, but
when you write it in base 36 it becomes prime.!
43,274,246_{10} = PRIME_{36}

David
Friday, October 30, 2015
Mr. B's Brain0 Quiz # 31
10/30/2015
When a problem comes along,
study it until you are completely knowledgeable.
Then find that weak spot,
break the problem apart, and the rest will be easy.
Mr.
B’s “BRAINO” Quiz
Guaranteed
to unclog clogged up brains!
NAME:__________________ DATE:____________
PERIOD:___
Directions: Carefully read and answer the
following questions. Print your answer
clearly in the box next to each question.
Questions are worth one million nano bonus points each. However, if you get all five correct, we will
double the points for a total of ten million nano bonus points.
1.


A man goes to
work. He knows that if he tears his
suit he will die. What does he do for
a living?




2.


Mrs. Chambers is
buying Christmas presents for her 7 children to give each other. If each child gives one present to each of
the others, how many presents will Mrs. Chambers have to buy?




3.


After Christmas,
two students are comparing the number of candy canes they have. “If I give you one of my candy canes, you
will have twice as many as I have; but if you give me one of your candy
canes, we will have the same number.”
How many candy canes do the students have?




4.


Write the number
twelve thousand twelve hundred twelve in simplest form.




5.


Mr. Smith was born
in St. Louis, during a leap year, on a date that is not divisible by 2, 3, or
5, during a month that does not have the letter “e” or “i” in it’s name. On what day will he become one year older?

When a problem comes along,
study it until you are completely knowledgeable.
Then find that weak spot,
break the problem apart, and the rest will be easy.
 Norman Vincent Peale
WARNING: For Educational
Purposes Only.
Please do not attempt to answer any of these questions
unless you are willing to learn something.
Please do not attempt to answer any of these questions
unless you are willing to learn something.
Answers will be
posted on Monday.
David
Thursday, October 29, 2015
Sam Loyd's 1415 Puzzle: Part III
10/29/2015
SOLUTIONS:
Figure 1: Sam explains that the "real trick" of the puzzle
could only be performed by changing the 9 into a 6, and the 6 into a 9, by
turning them upside down during the manipulation of the blocks. And I know this problem can be solved. The reason the original puzzle with the 14 and 15 reversed cannot be worked is because the reversal changes the parity of the puzzle. Flipping the 6 and the 9 to make them look like the 9 and 6 is the same as switching the 9 and the 6. This changes the parity again, returning the puzzle to a solvable state.
Starting from original position figure 2
may be reached in 44 plays as follows: 14, 11, 12, 8, 7, 6, 10, 12, 8, 7, 4, 3,
6, 4,7, 14, 11, 15, 13, 9, 12, 8, 4, 10, 8, 4, 14, 11, 15, 13, 9, 12, 4, 8, 5,
4, 8, 9, 13, 14, 10, 6, 2 and 1. (Moving the blank space to to top left rather than the bottom right, and then moving all the numbers to show them in the correct order leave every block in a different position  except for the 14  and ends up changing the parity and makes the parity match the beginning state of thee puzzle).
Figure 3 may be reached in 39 plays: 14,
15, 10, 6, 7, 11, 15, 10, 13, 9, 5, 1, 2, 3, 4, 8, 12 15, 10, 13, , 5, 1, 2, 3,
4, 8, 12, 15, 14, 13, 9, 5, 1, 2, 3, 4, 8, and 12. (Turning the puzzle a quarter of a turn changes the parity of the puzzle again, and allows a puzzlist to place all of the numbers in the correct order with the blank space back in the bottom right corner).
Figure 4: To produce a magic square adding
30 the following is best: 12, 8, 4, 3, 2, 6, 10, 9, 13, 15, 14, 12, 8, 4, 7,
10, 9, 14, 12, 8, 4, 7, 10, 9, 6, 2, 3, 10, 9, 6, 5, 1, 2, 3, 6, 5, 3, 2, 1,
13, 14, 3, 2, 1, 13, 14, 3, 12, 15, and 3. (The solution is a magic square with the same parity as the starting position of the puzzle with the 14 and 15 swapped. The the beginning arrangement and the final arrangement have the same parity this solution is possible. Now all you have to do is figure out which magic square arrangement to use. There are 7040 possible magic square arrangement and 3520 have the right parity to make this problem solvable.)
This problem had more meat on it than I thought.
I recommend downloading a copy of Sam Loyd's book from the internet.
I also recommend reading Jerry Slocum's book "The 15 Puzzle".
I recommend downloading a copy of Sam Loyd's book from the internet.
I also recommend reading Jerry Slocum's book "The 15 Puzzle".
David
Wednesday, October 28, 2015
Sam Loyd's 1415 Puzzle  Part II
10/28/2015
Now for the truth.
Sam did not invent the 15 puzzle. Sam did not have anything to do with the 15 puzzle craze that started late in 1879. And he did not start the craze about solving the puzzle with the 14 and 15 reversed.
The puzzle you saw yesterday came out in 1914, 3 years after Sam Loyd died. His son published a lot of the material that Sam had left in his papers, so I don't know for sure how much of what is said about the puzzle is true  though I do know quite a bit of it is false.
The $1,000 prize was never claimed, because it is impossible to solve with the 14 and 15 reversed (or any two numbers for that matter). So it was a safe bet. On that Sam knew he would never have to pay out.
However, it is an interesting puzzle. And it is worth studying.
Now for the truth.
Sam did not invent the 15 puzzle. Sam did not have anything to do with the 15 puzzle craze that started late in 1879. And he did not start the craze about solving the puzzle with the 14 and 15 reversed.
The puzzle you saw yesterday came out in 1914, 3 years after Sam Loyd died. His son published a lot of the material that Sam had left in his papers, so I don't know for sure how much of what is said about the puzzle is true  though I do know quite a bit of it is false.
The $1,000 prize was never claimed, because it is impossible to solve with the 14 and 15 reversed (or any two numbers for that matter). So it was a safe bet. On that Sam knew he would never have to pay out.
However, it is an interesting puzzle. And it is worth studying.
OLD AND NEW PROBLEMS: Sam suggests there is
a trick that will allow the original problem (with the 14 and 15 reversed) to
be solved. He also suggest three new
problems developed from the original puzzle which are worth sharing:
The Original Problem: Sam seems to hint
that there is a trick that will allow the original problem to be solved. “The real trick of the puzzle could only be
performed by changing the 9 into a 6, and the 6 into a 9, by turning them
upside down during the manipulation of the blocks.”
1

2

3

4

1

2

3

4


5

9

7

8

>

5

6

7

8

6

10

11

12

9

10

11

12


13

15

14

13

14

15

Figure 1:
Second Problem: Start again with the blocks
as in figure 1 and move them so as to get the numbers in regular order, but
with the vacant square at the upper left hand corner instead of lower
righthand corner; see figure 2.
1

2

3

4

1

2

3


5

6

7

8

>

4

5

6

7

9

10

11

12

8

9

10

11


13

15

14

12

13

14

15

Figure 2:
Third Problem: Start with figure 1, turn
the box a quarter way round and so move the blocks that they will rest as in
figure 3.
13

9

5

1

>

1

2

3

4

15

10

6

2

5

6

7

8


14

11

7

3

9

10

11

12


12

8

4

13

14

15

Figure 3
Fourth Problem: This is to move the pieces
about until they form a “magic square,” so that the numbers will add up thirty
in 10 different directions (4 rows, 4 columns and the 2 main diagonals  but he does not say what the magic square look like or where the blank space goes). See Figure
4.
1

2

3

4

13

1

6

10


5

6

7

8

>

14

2

5

9

9

10

11

12

12

11

7


13

15

14

3

15

8

4

Figure 4:
Tomorrow we will see if any of these problems can be solved or if Sam is hoaxing us one more time.
David
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