FUZZY
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On 10/23/2016 at 4:12 PM, araver said:
There are 3+4+5+4+3 = 19 huts so 19 ages of the eldest sons in each hut.
Sum of all ages is 5 times the sum of any row which is twice the age of the oldest sun in the village.
There are no gaps and no two are the same so the sum of the 19 ages.
Hence:
(i) S = 10*X
(ii) S = Y + (Y+1) + .... +(Y+18)
(iii) X = Y+18Rewriting (ii) as a consecutive sum:
S = 19*(Y-1) + (1+....+19) = 19*(Y-1) + 19*20/2
and using (i) and (iii):
10*X = 10*(Y-1)+10*19 = S = 19*(Y-1) + 19*10Therefore 9*(Y-1) = 0, so Y=1 and X = 19.
Did not use the fact that they are not of voting age (which indeed narrows the possibilities for Y to be either 1 or 2).
when you say the sum of all ages do you mean the sum of the ages of all the eldest sons or sum of the ages of all the male children in the village?
what is Y?
I really like the puzzle. But I am neither able to completely understand the question or the solution. Can you throw some light on both?
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On 10/22/2016 at 5:20 PM, bonanova said:
With a nod to jasen's recent and interesting puzzle,
A traveler happened upon a village of huts, laid out as the circles in the picture below. The village's mayor explained to the traveler that the family living in each of the huts had an eldest son whose age was unique within the village. (No two eldest sons had the same age.) How interesting, replied the traveler. Tell me this: of all the male children here, what is the age of the very oldest?
The mayor thought for a moment and replied, well I guess I could tell you that none of them are yet of voting age (21), and I guess you might be interested to hear that there are no gaps in their collective ages. But all of that wouldn't be enough information. I think it would be better for you to just knock on all the doors and ask.
I don't have time for that, replied the traveler, and I'm really not that interested. Well, here's an interesting thing about our village, replied the mayor. You may have noticed, our huts are laid out so that many rows of 3, 4, or 5 huts cut across the entire village. Just ask at the huts along any of those rows. Add the ages that you hear, and divide the sum by two. That way you will learn the age of the oldest son in the village.
And now, without knocking on any doors, you can learn it too.
what is meant by " there are no gaps in their collective ages"? Does it mean that the total age of all the male children in every hut is the same number?
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On 10/12/2016 at 3:55 PM, bonanova said:
First thought
2(1000/25) = 240. About 1.1 trillion>
Edit: Well, that's the number born in the 40th generation.
Adding all the generations gives double that amount.The first gen (children) are 5 yrs from now as per the question. So, the 25 yrs interval starts from the 2nd gen (grandkids). 1000 yrs from now .....would not it be from 1st gen onwards . Hmm... how 2^40 ? would not it be < 2^40? Not that it would make a lot of difference, it would still be in trillions, of course! Jus bein' curious .........that's all.
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On 10/13/2016 at 2:30 AM, BMAD said:
Suppose there is a hat that contains the numbers 1,2,3,4, and 5. You seek to find the three. You blindly reach into the hat pulling out a number. If it is wrong, then without replacement you reach for another. When you pull out the three you stop. Each pick before three is subtracted from 0 with your final pick (the three) being added to that total. What is your expected value?
You could get 3 in the first, second, third, fourth or fifth draws.
3 in the first draw: One number (3). 1 way
In the second draw means _ 3 : for the number drawn before 3, it is 4C1 ways. Then, for 3, it is 1 way. So, 4*1= 4 numbers have 3 in second place. (13,23,43,53)
In the third draw means _ _ 3 : for the two numbers drawn before 3, it is 4C2. The two numbers drawn can arrange themselves in 2! ways. Next, 3 is drawn (1 way).
4C2*2!*1= 12 numbers have 3 in third place.
In the fourth draw means _ _ _ 3 : for the three numbers drawn before 3, it is 4C3. The 3 numbers can arrange themselves in 3! Ways. Finally, 3 is drawn.
4C3* 3!* 1= 24 numbers have 3 in fourth place.
In the fifth draw means _ _ _ _ 3: for the four numbers drawn before 3, it is 4C4. The 4 numbers can arrange themselves in 4! Ways. Finally, 3 is drawn.
4C4* 4!*1= 24 numbers have 3 in fifth place.
Totally, 1+4+12+24+24 = 65 numbers have 3 in them.
First draw value = (1/65)*3 = 3/65
Second draw (13 or 23 or 43 or 53) value for the four numbers
= (1/65)*(0-1+3) + (1/65)*(0-2+3) + (1/65)*(0-4+3) + (1/65)*(0-5+3)
= 1/65 [–(1+2+4+5)+(3*4)]= 0
Third draw value for the twelve numbers =-36/65
Fourth draw for the 24 numbers = -144/65
Fifth draw = -216/65
Total expected value= (3/65)+0+(-36/65)+(-144/65)+(-216/65)= -393/65
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3 hours ago, Buddyboy3000 said:
@FUZZY Does the 1000 years start after your children has two children, or from when the question starts?
from when the question starts.
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23 hours ago, Buddyboy3000 said:
The family was able to go on the picnic.
When it says the certain day was yesterday, and the day before yesterday is yesterday, they are equal to each other. You are only confused by all the words. So, the certain day is the day before yesterday, which is two days ago. Then, the family will go on the picnic the day after tomorrow from the certain day. The family will go on the picnic today. If it rained the day before yesterday, then the family will be able to go on the picnic today.
how do you decide the days are equal?
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Suppose the certain day is Sunday, it would've been yesterday on Monday. The day before yesterday with respect to Monday is Saturday. Saturday would've been yesterday on Sunday. So, Sunday was yesterday (on Monday) when Saturday was yesterday (on Sunday)?? The first sentence does not make sense.
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A certain day was yesterday when the day before yesterday was yesterday . On that certain day Kim's parents said "we'll go on a picnic the day after tomorrow if it doesn't rain". It rained the day before yesterday but not since.
Did the family go on the picnic? -
I am not smart enough to understand the procedure explained above. But I did look up Hungarian algorithm.
I too got 18 as TC. Sorry for the earlier mistake.
1 2 3 4 5 A 8 3 5 4 3 B 2 6 9 4 7 C 6 1 8 4 3 D 5 7 9 8 8 E 5 7 9 4 3 After row minima's subtracted 1 2 3 4 5 A 5 0 2 1 0 B 0 4 5 2 5 C 5 0 7 3 2 D 0 2 4 3 3 E 2 4 6 1 0 After column minima's subtracted 1 2 3 4 5 A 5 0 0 0 0 B 0 4 3 1 5 C 5 0 5 2 2 D 0 2 2 2 3 E 2 4 4 0 0 After assigning zeroes 1 2 3 4 5 A 5 0 0 0 0 B 0 4 3 1 5 C 5 0 5 2 2 D 0 2 2 2 3 E 2 4 4 0 0 Marking unassigned rows 1 2 3 4 5 A 5 0 0 0 0 B 0 4 3 1 5 C 5 0 5 2 2 D 0 2 2 2 3 x E 2 4 4 0 0 Marking columns corresponding to the marked row 1 2 3 4 5 A 5 0 0 0 0 B 0 4 3 1 5 C 5 0 5 2 2 D 0 2 2 2 3 x E 2 4 4 0 0 x Marking row having assigned zero corresponding to the marked column 1 2 3 4 5 A 5 0 0 0 0 B 0 4 3 1 5 x C 5 0 5 2 2 D 0 2 2 2 3 x E 2 4 4 0 0 x Drawing lines through unmarked rows and marked columns 1 2 3 4 5 A 5 0 0 0 0 B 0 4 5 1 5 x C 5 0 5 2 2 D 0 2 2 2 3 x E 2 4 4 0 0 x
No of lines not equal to 5. So, we work further on the table values. After subtraction of minimum value among uncovered cells from the latter and adding the min. value at the intersection of the lines 1 2 3 4 5 A 6 0 0 0 0 B 0 3 4 0 4 x C 6 0 5 2 2 D 0 1 1 1 2 x E 3 4 4 0 0 x Assigning zeroes 1 2 3 4 5 A 6 0 0 0 0 B 0 3 4 0 4 C 6 0 5 2 2 D 0 1 1 1 2 E 3 4 4 0 0 A3 B4 C2 D1 E5 TC 5 4 1 5 3 18 -
how to arrive at the answer?
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Very few people bother to determine what an amazing potential for overpopulation they have. Let us consider yours. If you and your wife have a meager two children, it would seem you are not adding much to the world. But let's suppose that in 5 years each of your children has two children. In another 25 years, each grandchild has two kids. In another 25 years, each great-grandchild has two offspring-and so on for a thousand years. How many descendants would you have?
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Given the sum and the product of the digits of a 7-digit even number with distinct digits that is divisible by the product, is it possible to figure out the number?
Eg: Given : sum of the digits=36, product of the digits=18144. Also given: the 7 digit number with distinct digits is divisible by the product 18144.
how do you figure out the 7-digit even number as 1687392?
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once you know the sum as 36 and the product as 18144, how do you figure out the number as 1687392?
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A3 B1 C2 D4 E5
5+2+1+4+3 =15
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Samantha=39 , Jacob=18
Bonanova got every other number right. kudos!
Ellery=21 Jesse=18 Evelyn=12 Delphine=9
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Samantha had three children : Ellery, Jacob and Jesse.
Their combined ages totaled half of her age.
Five years later, during which time Evelyn was born, Samatha's age equalled the total of all her children's ages.
Ten more years have now passed.
Delphine was born during that time.
At that event Ellery was as old as Jessy and Evelyn together.
The combined ages of all the children are now double Samantha's age, which is only equal to that of Ellery and Jacob together.
Ellery's age also equals that of the two daughters.
Question : How old are they?How to arrive at the answer? During the 5 yrs, if Evelyn is born, it could be any number < 5, isn't it? I am clueless as to how to solve this puzzle.
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letterhead
fountain head
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So you have to assume c=1 and try 2 to 9 for O? If it is deduced that c=2, then again try all other values for O?
isn't there a shorter and more efficient way of solving?
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dead sea?
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one's own reflection.
your eye.
But, alive.
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but how is the answer arrived at without using computer?
what is the logic used in arriving at the answers?
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Spoiler
mail/greeting card
sausage
can't figure out the matchstick thing
holding hands
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the big ben
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The eldest son
in New Logic/Math Puzzles
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