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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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