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NOVELTIES
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- THE
MATHEMATICAL
CALENDAR
CONSIDER
ANY
DATE AFTER 1582 (ADOPTION OF
GREGORIAN CALENDAR)
LET : N=DAY OF THE MONTH, M=MONTH NUMBER
(starting with march=1),
C=HUNDREDS PART OF THE YEAR, Y=OTHER PART
OF THE YEAR,
L=1 IF LEAP YR, 0 OTHERWISE, [X]=GREATEST
INTEGER LESS THAN OR EQUAL TO
X.
THEN
THE
WEEK DAY IS :
d =
N+[2.6M-0.2]+Y+[Y/4]+[C/4]-2C-(1+L)[M/11]
(mod7)
(Sunday = 0, Monday =1, etc) ( mod 7 =
remainder when divided by 7 )
EXAMPLE
: Oct 18 1998
N=18, M=8, C=19, Y=98, L=0
d =
N+[2.6M-0.2]+Y+[Y/4]+[C/4]-2C-(1+L)[M/11]
(mod 7)
d =
18+[20.6]+98+[98/4]+[19/4]-38-(1)[8/11]
(mod 7)
d = 18+20+98+24+4-38-0 (mod 7)
d = 126 (mod 7)
d = 0 ------> Sunday
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- THE
MISSING
DOLLAR PROBLEM
Three
people
have dined at a restaurant and received a
total bill for $30. They
agree to split the amount equally and pay
$10 each. The waiter hands
the bill and the $30 to the manager, who
realizes there's been a
mistake and the correct charge should be
only $25. He gives the waiter
five $1 bills to return to the customers,
with the restaurant's
apologies. However, the waiter is
dishonest. He pockets $2, and gives
back only $3 to the customers. So, each of
the three customers has paid
$9 and the waiter has stolen $2 making a
total of $29. But the original
bill was for $30. Where is the missing
dollar ?
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A dozen, a gross, and a score
Plus three times the square root of four
Divided by seven
Plus five times eleven
Is nine squared and not a bit more.
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- ARISTOTLE'S
WHEEL
PARADOX
THE OUTER CIRCLE TURNS ONCE WHEN
GOING FROM A TO B, AS DOES
THE INNER CIRCLE WHEN GOING FROM C TO D.
HOW CAN THIS BE?
AB IS THE SAME LENGTH AS CD, BUT THE
CIRCLES ARE A DIFFERENT SIZE.
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If
a
number is chosen at random from a large
table of data or statistics,
the chance that the first digit is 1 is
about 30.1%, that the first
digit is 2 is about 17.6%, that it
is three is 12.4%, . . . . . ,
that it is 9 is 4.5%. In fact,
the probability the first
digit is d is log(1+1/d).
Benford tested thousands of different
collections of data, including
the surface area of 335 rivers, specific
heats and molecular weights of
thousands of chemicals, baseball
statistics, and addresses chosen
at random.
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Given
a
sandwich in which bread, ham, and cheese
(three finite volumes) are
mixed up, in any way at all, there is
always a flat slice of a knife (a
plane) that bisects each of the ham,
bread, and cheese. In other words,
however messed up the sandwich – even if
it's been in a blender – you
can always slice through it in such a way
that the two halves have
exactly equal amounts, by volume, of the
three ingredients.
Pick
a
door. Behind one of them is a NEW SAIL
BOAT, and behind the other two
are goats. The Monty hall problem as
explained by Stedwick
.
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NOTES
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