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Posts posted by jasen
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1. Easy Puzzle
Put numbers 1,2,3,5,6,7 to each point, so :
I. The sum of points in every circle are equal, (A+E+F+C = B+E+D+C = A+D+F+B)
II. A+B+C = D+E+F
There are 2 solutions.
2 .Harder Puzzle
Find the value of x, then put number 2,3,4,9,12,x to each point, so :
I. The product of points in every circle are equal, (A*E*F*C = B*E*D*C = A*D*F*B)
II. A*B*C = D*E*F
There are 2 solutions.
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If you find the problem "magic cube 2x2x2" is too easy, try this.
in a 2x2x2 cube
Put number 1, 2,3,4,5,6,8, and 10 to every small cube,
so the product of each side of the cube are equal.A*B*C*D = E*F*G*H = B*D*F*H = A*C*E*G = A*B*E*F = C*D*G*H
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There is 1 number can't written like that. The number is 0.
So the statement is wrong.
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We are 3 brothers, we like multiplying our ages each other.
m years later the result is m times than today
n years later the result is also n times than today (m and n is not equal)
we find it is interesting.
How old we are?
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Spoiler
Possibilities1 L:Natalia R:Empty = true stat : 1,3
2 L:Lion R:Empty = True Stat : 2
3 L:Empty R:Natalia = True Stat : 3
4 L:Empty R:Lion = True Stat : 2,4
5 L:Natalia R:Lion = True stat : 1,2,3,4
6 L:Lion R:Natalia = true stat : 3,4
7 L:Lion R:Lion = true stat : 2,4
8 L:Empty R:Empty = True stat : nonePossibility number 2 and 3 are quailified
So choose right door, since you may find Natalia or empty room, at least you will not eaten by lion.
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Similar, but a bit harder puzzle :
A father left 102 goats to his three sons. He promised 1/2 of the goats to the oldest son, 1/5 to the middle son, and 1/13 to the youngest son (maybe step son, just get a bit heritage ). But the sons could not divide 102 evenly. Then they call their uncle, after thinking a while the uncle come with a solution, and as thanks they give the uncle 1 goat, what is the uncle solution ?
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A father left 45 goats to his three sons. He promised 1/2 of the goats to the oldest son, 1/4 to the middle son, and 1/6 to the youngest son. But the sons could not divide 45 evenly. Then they call their uncle, after thinking a while the uncle come with a solution, and as thanks they give the uncle 1 goat, what is the uncle solution ?
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16,363636 minutes
SpoilerThe hour and minute hands are superimposed only when their angle is the same.
in 12 hour, hour and minute hands superimposed 11 times.so time between 2 superimposed is 12/11 hour = 65.454545... minutes
greatest negative (receding) speed when minute and hour hand form -90 deg.
greatest positive (approaching) speed when minute and hour hand form 0 deg.because 1 circle is 360 deg, so :
time from negative to positive speed speed is(12/11)/4 hour = 16,363636 minutes
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If we make sequences of the numbers, we get this pattern :
Month
0
1
2
3
4
5
6
7
8
9
10
11
12
sequence
1
1
3
5
11
21
43
85
171
341
683
1365
2731
This pattern is called Jacobsthal numbers, Like Fibonacci numbers, sequence follow the rule :
The first two numbers in the sequence are 0 and 1, then each following number is found by adding the number before it to twice the number before that.
In math expression : Jn = 2*J(n-2) + J(n-1)
The formula to find the x-th number is Jn = ((2n)-(-1)n) / 3
So after 1 year there are J13 = (213 + 1) / 3 = 2731 pairs of rabbits
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@logophobic : Right answer, with very clear explanation.
Can you see any pattern there ?
Can we generalize the problem, such as if a pair begets 3,4, or any number new pairs ....
or any pairs begets any number new pairs,....clue : generalization of fibbonacci number.
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A man put a pair of rabbits in a place surrounded on all sides by a wall. How many pairs of rabbits can be produced from that pair in a year if it is supposed that every month each pair begets two new pair which from the second month on becomes productive?
This problem states several important factors:
• Rabbits take one month to grow up
• After they have matured (for one month) it takes a pair of rabbits one more month to produce their first 2 pairs of newly born rabbits.
• We assume that rabbits never die
• We assume that whenever a new pair of rabbits is produced, it is always a male and a female
• We assume that these rabbits live in ideal conditions
• The problem begins with just one pair of newly born rabbits (a male and a female). Given allthis information, how many pairs of rabbits will there be in one year (12 months)?
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16 hours ago, DejMar said:
Given that the first mathematician was unable to know the ages of the four children, the sum of the ages occurs multiple times for different sets of integer ages where the sum of the reciprocals equals the multiplicative identity. This only occurs where the sum is 22 (which must be the age of the first mathematician's sister). These values are: (8,8,4,2), (6,6,6,2), (12,4,3,3).
Given the oldest two are twins reduces this to two possibilities: (8,8,4,2) and (6,6,6,2). The first set is likely the intended answer, though the second can not be dismissed without consideration. (It is a fact that a child can be born less than a year after an older sibling, thus for a short period they can have the same age). Still, considering that a single solution is sought, the ages themselves are to be considered the compose the twins (as opposed to the children), and thus (8,8,4,2) would be the answer.
Right Answer, but you make a bit miscalculation. sum of 6,6,6,2 is 20, not 22.
Spoiler1st Clue : Sum of their multiplicative inverse is 1
All age possibility are
4th child
3rd child
2nd child
1st child
sum ages
2
3
7
42
54
2
3
8
24
37
2
3
9
18
32
2
3
10
15
30
2
3
12
12
29
2
4
5
20
31
2
4
6
12
24
2
4
8
8
22
2
5
5
10
22
2
6
6
6
20
3
3
4
12
22
3
3
6
6
18
3
4
4
6
17
4
4
4
4
16
2nd Clue : Sum of their age is your sister age
Because “mat 1” said that knowing the total (from the age of her sister) did not help, we know that knowing the sum of the ages does not give a definitive answer; thus, there must be more than one solution with the same total.
Only three sets of possible ages add up to the same totals:
2
4
8
8
22
2
5
5
10
22
3
3
4
12
22
3rd clue : 2 oldest are twin
“mat 2” concludes that the correct solution is 8,8,4,2.
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2 mathematicians meet at their school reunion.
mat 1 : Hey old friend, I heard you have 4 child, How old are they?
mat 2 : Sum of their multiplicative inverse is 1, and sum of their age is your sister age.
mat 1 : But, I still don't know their ages.
mat 2 : 2 oldest are twin.
mat 1 : ok, I know know. -
50 minutes ago, Rob_G said:
Hidden Content
Spoiler1/3 + 1/3 + 1/6 + 1/6 = 1
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2 mathematicians meet at their school reunion.
mat 1 : Hey old friend, I heard you have 2 pair identical twins, How old are they?
mat 2 : Sum of their multiplicative inverse is 1.
mat 1 : Ok, I know. -
Ok I was wrong, but I know the number is 2x3x2x5x7x2x3x11x13x19x23x5x3x29 = 68502634200
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if the number is 65520, two consecutive numbers are 22 & 23
two students spoke wrongly are 21th and 22th studentif the number is 166320, two consecutive numbers are 25 & 26
two students spoke wrongly are 24th and 25th student- 1
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statistically it is uniform.
1000 places of Pi
3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989
1=116, 2=103, 3=103, 4=93, 5=97, 6=94,7=95, 8=101, 9=106, 0=93 total digits 1001
Also for base2, 100008 first digits, have 50052 1
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Advantage for the 1st player, because he can keep tracking and control remaining dashes.
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Thank you bubbled, I understand now.
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the answer is same like horse racing puzzle.
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I think bubbled is right,
"Sorting like" strategy, without "real sorting" will not work.I have proposed this "Sorting like" strategy, exactly like plasmid.
How about this :
Hidden Content
but then I realize it was not working. I've tested it using dev-C++.
the chance to find your name is only 50%, the same chance with random way opening boxes.in my program : box numbered from 0 to 9, and your number is random from 0 to 9. You can only open 5 boxes to find your name.
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computer simulation proved you are right !!
I think we can go a little higher
Hidden Content
three circles
in New Logic/Math Puzzles
Posted
@DejMar
1st question
1 5 6 3 2 7 = 1 6 5 2 3 6 = 5 1 6 7 2 3 = 5 6 1 2 7 3 = 6 1 5 7 3 2 = 6 5 1 3 7 2 1st solution
2 3 7 5 1 6 = 2 7 3 1 5 6 = 3 2 7 6 1 5 = 3 7 2 1 6 5 = 7 2 3 6 5 1 = 7 3 2 5 6 1 2nd solution
2nd question
2 9 12 4 3 18 = 2 12 9 3 4 18 = 9 2 12 18 3 4 = 9 12 2 3 18 4 = 12 2 9 18 4 3 = 12 9 2 4 18 3 = 1st solution
3 4 18 9 2 12 = 3 18 4 2 9 12 = 4 3 18 12 2 9 = 4 18 3 2 12 9 = 18 3 4 12 9 2 = 18 4 3 9 12 2 = 2nd solution
Just Rotation and reflection of the circle.