superprismatic
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No need to spoiler.
If it has made a complete revolution, (and it has) it's facing the same "way" after as before.
Maybe I'm missing something.
I think you have to determine what the coins are to determine which way the head is facing. If the rotating coin were a US cent, then the head faces to the viewer's right. Nearly all other US coins (except for very recent mintings) have heads that face to the viewer's left. I suppose that you have to use the circumferences of the coins to find out which pair would satisfy the other conditions of the OP.
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a) 3
b) y = (-5/6)x3 + (15/2)x2 - (62/3)x + 21
c) No. But it does go through (1,7)
I tried solving for the coefficients of a quadratic using gaussian elimination
without success, as the system of linear equations was inconsistent. When I
did the same thing with the coefficients of a cubic, I found the system to be
consistent with the coefficients (-5/6), (15/2), (-62/3), and 21. Previously,I had erred in solving the cubic case and went on to the quartic with success.
The quartic I had in post #2 is the smallest degree monic polynomial that does
the job. Monic, however, was not a requirement.
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a) 4
b) y = x4 - (89/6)x3 + (157/2)x2 - (524/3)x +141
c) No. But it does go through (1,31)-
1
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In sequencing DNA, technicians break up long molecules into
smaller pieces because it is easier to sequence small segments
than long ones. They must then reconstruct the target molecule
by fitting the small segments together using overlaps as a guide.
The small segments are sequenced beginning at either end in a
random fashion. So, for example, the segment ATACAG may also
be sequenced as GACATA.
In this puzzle, I have chopped up many identical 50-long sequences
into pieces of lengths 5, 6, 7, and 8. 30 of these pieces are:
ATACAG
TGACAT
GTCTTA
GTCGAGA
AACGA
CAAGG
CAGTGTGA
GTGGTGT
CCGATGAC
AGACAA
TGTGA
ATACAGTG
CTGTG
ACATA
GTGTCG
TACAGT
GTCTTAG
CATAA
GGTGG
GACTCCAG
TGTGA
CCAGTG
TGTGACA
GTGACATA
TGTGGT
ATTCTGA
TGAATACA
TAGCCG
TGTGACCT
CCTCA
I have insured that these 30 pieces completely cover the 50-long
sequence with which I started. Can you find that sequence?
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1
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1
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1.25, 1.60, and 2.85.
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For part a: 2-5-1-6-3-7-4-8 with a distance of 51.253
For part b: 3-6-1-5-2-7-4-8-3 with a distance of 56.304-
1
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PVJBIVEA
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(googol+1)/2
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I assume a two-dimensional ladder:
If x is the acute angle formed between the ladder and the outside wall
of the 4 foot corridor, then the longest ladder to form it will be
(4/sin x)+(6/cos x). This is maximal at x=arctan((2/3)^(1/3)). The longest
ladder, therefore, has length approximately 14.047 feet. So, the 14 foot
ladder makes it with about .047 of a foot to spare. -
The probability of picking one of each color from bags 1&2 is 14/36.
The probability of picking one of each color from bags 1&3 is 18/36.
The probability of picking one of each color from bags 2&3 is 18/36.
So, by Bayes, the probability of having used bags 1&2 initially, given that one of each color was chosen is 14/50.
So, by Bayes, the probability of having used bags 1&3 initially, given that one of each color was chosen is 18/50.
So, by Bayes, the probability of having used bags 2&3 initially, given that one of each color was chosen is 18/50.
Therefore, the remaining bag is 3 with probability 14/50;
the remaining bag is 2 with probability 18/50;
the remaining bag is 1 with probability 18/50;
So, the probability of picking a white from the remaining bag is (14/50)*(1/2)+(18/50)*(2/3)+(18/50)*(5/6)
which equals .68 exactly. -
.68 precisely.
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4x10
249-1, here's why:The pairs must contain only odd numbers, so this brings us down to 10250 possible
pairs. Furthermore, each pair must contain only those odd numbers which are not
equal to 0 modulo 5. This brings us to (4/5)x10250 which equals 8x10249. But,
this is 2 too many because the sequence of odd numbers begins (modulo 5) with
1,3,2,4,1,3,2,4... but ends in 1,3 which is the middle of this 4-long cycle.
Which brings us to 8x249-2 pairs. But this is double what we want because
it counts each pair twice, counting the order of addition both ways. Therefore,
we need to divide this in two to get 4x10249-1 pairs.
Consider the case of 2 x 101. Answer: 4 x 100.
1 19
2 183 17
4 165 156 147 13
8 129 11
10 10Suppose I were to argue that 2x10250 constructs an integral number of similar groups of 10 sums, each of which gives back 6 rows to the mutually prime condition.
Which group contributes the (-1)?
It seems rather difficult to pull the wool over the Denizens!
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4x10249-1, here's why:
The pairs must contain only odd numbers, so this brings us down to 10250 possible
pairs. Furthermore, each pair must contain only those odd numbers which are not
equal to 0 modulo 5. This brings us to (4/5)x10250 which equals 8x10249. But,
this is 2 too many because the sequence of odd numbers begins (modulo 5) with
1,3,2,4,1,3,2,4... but ends in 1,3 which is the middle of this 4-long cycle.
Which brings us to 8x249-2 pairs. But this is double what we want because
it counts each pair twice, counting the order of addition both ways. Therefore,
we need to divide this in two to get 4x10249-1 pairs. -
I solved it with by exhaustion over all possible paths of moves less than 38.
I saw that Martin Gardner's column of 1975 had a 38-move solution, so I started there.
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I found a novice programmer mistake in my code which computed the values in
my post #14. The value I get now is the same that bonanova and BMAD get.
I'm relieved that my approach agrees with the concise generating function one.
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I found a 36 long solution:
URDL ULDR DLUR DRUU LDLU RDRU LDLU RRDL LDRU -
My approach was to exhaust over all possible outcomes.
Lets look at what can possibly be chosen from a particular
value of card and look at what can happen with the four
7s in the deck. With Probability 1/16, no 7s are chosen
which contributes 0 to the running sum; with probability
4/16, one 7 is chosen which contributes 7 to the running
sum; with probability 6/16, two 7s are chosen which
contributes 1 (mod 13) to the running sum; with probability
4/16, three 7s are chosen which contributes 8 (mod 13)
to the running sum; with probability 1/16, four 7s are
chosen which contributes 2 (mod 13) to the running sum.
So, for the 7s in the deck we have 5 possible contributions
to the mod 13 sum with their respective probabilities.
We can do this with all 13 sets of card values. So, we can
compute the contributions of all 513 possible ways of
chosing cards from each card value. Now 513 is about 1.22
billion ways. We also know the probabilities involved, so
we can compute the probability of each way of forming any
particular mod 13 value. Furthemore, we can do the
calculations with all integer arithmetic by using 1,4,6,4,1
instead of 1/16, 4/16, 6/16, 4/16, 1/16. When we do this,
all the mod 13 counts we get had better add up to 252.
This is a check on the program. I wrote the program and it
takes about 5 seconds to run on my Intel Core7 machine.
The mod 13 sums and counts are as follows:0 346442068663808 1 346425832982400 2 346393927579136 3 346477138939392 4 346368483871232 5 346474428126848 6 346463204652160 7 346402730352000 8 346453603433856 9 346431886650368 10 346395126293504 11 346461321451008 12 346409874374784 4503599627370496 Total which is 2^52
So, the probability of a sum of 0 (mod 13) is
346442068663808/4503599627370496
which is approximately 1/12.99957492097 -
The exact probability of the sum being 0 mod 13 is
676644665359/8796093022208 which is approximately 1/12.99957492097 -
The exact probability of the sum being 0 mod 13 is
1353275446583/17592186044416 which is approximately 1/12.99970829208 -
17710 Female ancestors and 10945 male ancestors for a total of 28655.
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669640625 ( 1 is the third funny number )
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Whoops! I overlooked the word "consecutive".
~.5386
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13/49
Penny Triangle
in New Logic/Math Puzzles
Posted