The worlds first riddle!

It's much simpler than that. There are N! arrangements (assuming unique flavors), and you're picking one of them. P(all correct) = 1/N!.

Rap: Your counting method is wrong. For each N, you show only one way to get them all wrong. In fact, as N grows, the number of ways grows. For example, with three cans, there are two ways to get them all wrong:
Actual: A B C
Wrong: B C A
Wrong: C A B

With four cans, there are nine ways to get them all wrong.

Markr I see my error and see that the possibility of getting all right in a random pick out of N objects is 1/N!.

And when I look further at four and the number of permutations and the number of errors I count

There are 4! permutations of which
4 correct is 0 permutations & 0 is an even number
2 correct is a single permutations & 1 is an odd number
1 correct is 2 permutations & 2 is an even number
0 correct is 3 permutations & 3 is an odd number

now even permutations + odd permutations=4!
&
even permutations=odd permutations=4!/2
even = 4 correct+1 correct & 4 correct=1
so 1 correct=11 permutations
& if 0 correct is 9 permutations then 2 correct is 3 permutations.

So the odds for 4 are
4 correct=1/24
2 correct=3/24
1 correct=11/24
0 correct=9/24

But when I expand to 5
I get
5 correct (0 permutations)
3 correct (1 permutation)
2 correct (2 permutations)
1 correct (3 permutations)
0 correct (4 permutations)
Now I know that even permutations are 0 & 2 & 4 permutations and odd permutations are 1 & 3 permutations.
Moreover the even permutations are equal to the odd permutations which is equal to 5!/2 or 60. Where I keep getting into problems is counting the number of permutations in each case. There is (obviously) only one 0 permutation (all 5 right)

Moreover since any single permutation out of 5 (3 right) is switching AB, AC, AD, AE (4), BC, BD, BE (3), CD, CE (2). DE (1) is 10 (4+3+2+1) Note this is the fourth triangular number. So the number of single is 10 and three permutations is 50 (60-10) which is 1 right.
So the possibility is
5 right is 1/120
3 right is 10/120
1 right is 50/120 Note this is a special and not a general case
but how do you determine the number of 2 and 0 right permutations?

Suggestions are being solicited.

Rap c∫;?/

Markr....

Wouldn't you also be able to chose ACB and BAC? You must mach them in any possible order right?

Narkr was just listing the 'all wrong' permutation in a set of three to me. If the 'right' permutation was
ABC
the permutations ACB, BAC, & CBA would have 1 in ther right position

and BCA and CAB would have none in the right position.

In toto though if there would be 6 (3!) potential permutations.

By even an odd permutations I am talking about a pair switch, e.g. if you start with ABC and switch A and C you get CBA and you have one in the right place. Since 1 is an odd number this would be denoted a odd permutation.

To get to an even permutation pick any pair of CBA and switch them (say BA) to get CAB now you have none of them in the correct position (none right).

Calling 0 an even number, then for three items all right and all wrong are even permutations and
one right is an odd permutation. And for all 6 permutations 3 are odd and 3 are right.

This is interestingly a property that can be generalized for sets of N items. There are N! permutations and N!/2 of them are even and N!/2 of them are odd.

Rap

Mark:

BIRTHDAYS
Assuming 365 days in a year, 23 and 41.

JUICE CANS
2: 1/2
3: 1/6
4: 1/24
10: 1/10!
100: 1/100!
N: 1/N! WOW!

REVERSED DIGITS = 90

ARROWS = 9

ABC = 5/8

TV = 2


Rap:
Birthday Odds
This is a reversal problem--first you have to figure the probability of none of them share a birthday and subtracting this probability from certainty. The result, one birthday (or more) shared
P(E)=50% E is about 22 Pretty
P(E)=90% E is about 40 ditto.
I've always heard it is a safe bet at 30. This is where the P(E) is about 72%

Juice Flavor
I think there something missing here Try, perhaps your forehead has an increasing gradient too? You could be right, but not this time. :wink:

Palindrome numbers between 100 and 1000
Including 101 and 999 there are 90

Whim and Drewdad's arrows
2A+B=17
A+2B=22
3B=27 B=9

a/c
a/b=3/4
b/c=5/6
a/c=a/b*b/c=3/4*5/6=15/24=5/8

T&V
T37V=2376
2376/88=27
T=2

Birthdays: Great answers!

One way to solve this is to turn the problem around and think about how likely it is for there to be NO matches in a group of a given size. If there is only one person a room there can be no shared birthdays since there is no one to share with. The probability of not having a match in this case is 1. Events that are certain are said to have a probability of 1. At the other extreme, with 367 people in the room, it is certain that there will be at least one shared birthday since there aren't enough birthdays to go around.

Now imagine that a second person walks into the room. The probability of that person not having the same birthday as the first occupant of the room is 365 / 366 or 0.997. There are 366 possible birthdays and only one of them is a match.

Now if the first two people in the room have different birthdays and a third person walks in, there are two days used up so the probability of the third person not sharing a birthday with either roommate is 364 / 366 and the probability of no sharing amongst the three of them is 1 * 365 / 366 * 364 / 366 = 0.992, which is still over 99%. So with 2 or 3 people in the room there is less than a 1% chance of a shared birthday.

You can continue to calculate the chances of not having a shared birthday for any number of people:

1 * 365 / 366 * 364 / 366 * 363 / 366 * 362 / 366 ...
Things change quickly as the number of people increases. With 10 people in the room there is a better than 10% chance of a match. When there are 23 people in a room the chance of a shared birthday is slightly greater than 50% and it rises above 90% with 41 people.


Juice: Mark knows it, and I learn more from the different direction of thinking. Welcome back Zippy, I had alerted the FBI and Interpol. How high did you go?

One way to think about this problem is to look at the number of possible ways there are to arrange the cans of juice in a row. If you have two cans, let's say orange juice and apple juice, there are two possible arrangements:

apple orange / orange apple

One of these will be the correct arrangement. Since there are two possible guesses, the probability of guessing correctly is 1 out of 2 or 1/2.

Let's add grape juice. There are now six possible arrangements:

apple orange grape / apple grape orange /orange apple grape /orange grape apple /grape apple orange /grape orange apple.

Again, only one of these is the correct order so your chance of guessing correctly is 1 out of 6 or 1/6.

If you add a fourth can you will find that the number of possible arrangements is 24 and, therefore the probability of a correct guess is 1/24.

In general, the number of possible arrangements of N objects is

N * (N -1) * ... * 1

This is called the factorial of a number and is written N!

2! = 2
3! = 6
4! = 24

As N increases, N! increases very quickly. If you had 10 cans of juice, there would be 10! Possible arrangements. 10! = 3,628,800 so your chances of guessing correctly are rather small. With 100 cans of juice there are approximately

93,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000
000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000 possible arrangements.

In general, the probability of correctly guessing all the flavors when you have N cans of juice is 1/N!


Try this series:

a) 1, 4, 9, 16, 25, ...


b) 1, 2, 6, 24, 120, 720, ...



You are traveling over the Pyrenees by air between two cities that are 1,600 kilometers apart. Your plane flies at a speed of 800 kilometers per hour. If there is no wind, your flying time is 2 hours each way. The round trip takes 4 hours.

But what if there is wind?

Let's say that on the way from City A to City B there is a headwind of 200 km/h. This means that your ground speed is only 600 km/h. But on the way back you have a tail wind of 200 km/h so your ground speed is 1,000 km/h.

How does the wind affect your round trip time? Is it longer, shorter, or the same as in calm conditions



In the target, ring A, ring B, and circle C have different point values. The sum of the point values of A and B is 23, of B and C is 33, and of A and C is 30.

What is the sum of the point values of A, B, and C


There are many three-digit numbers that are each divisible by 7 and also by 8 without remainder in each case. What is the largest of these three- digit numbers


The entire treasury of the Alpha club, consisting of $240, was to be divided into equal shares for each club member. When it was discovered that one member was not eligible for a share, the share of each of the remaining members increased by $1.

How many eligible members got a share of the $240


(1,1,8) is a triple of natural numbers which has a sum of 10. Consider (1,8,1) and (8,1,1) to be the same triple as (1,1,8). How many different triples of natural numbers have a sum of 10? Include (1,1,8) as one of your triples


When a certain number N is divided by 3, the result is the same as when N is decreased by 8. What is the number N


That should sort the men from the boys.

Thanks a bunch for the clarification. I actually wrote the permutations of 3 out like you....but then I decided to look at it in a silly way. I though about the permutations of the guess's instead. So if not shown the result of your first guess or the second while guessing all 3 cans at one time, I though one could guss as follows:
wrong,wrong,wrong
right,right,right
right,wrong,wrong
right,wrong,right
right,right,wrong
wrong,right,right
wrong,wrong,right
wrong,right,wrong

Thus one out of eight guess's would be correct.

Anyway, the trip was great! We started out at an altitude of about 2000m and made our way down to about 1500m. It was a first for me, but I am now hooked!

As far as the flight times go....they would be the same...rulling out that with wind, the plane would have to do a roundabout for takeoff to ensure that take off is with head wind and not tail wind....as planes don't like that sort of thing.

Working on your other problems now, but Markr will probably beat us all to 'em!

[size=7]SERIES
a) N^2
b) N!

FLYING
longer

TARGET
43

7 AND 8
952

ALPHA CLUB
15

TRIPLES
8

N
12[/size]

CANS

Rap,
The number of ways of getting exactly M correct out of N is:
C(N,M) * W(N-M)
where
C(n,r) is the usual combination of n things taken r at a time
and
W(r) is the number of ways of getting r out of r wrong

W(0) = 1 because it makes the math work out!
W(1) = 0 which is why you can't get all but one right.
W(2) = 1
W(3) = 2

Therefore,
4 right out of 4 = C(4,4) * W(0) = 1 * 1 = 1
3 right out of 4 = C(4,3) * W(1) = 4 * 0 = 0
2 right out of 4 = C(4,2) * W(2) = 6 * 1 = 6
1 right out of 4 = C(4,1) * W(3) = 4 * 2 = 8
Total = 15
Therefore,
0 right out of 4 = 4! - 15 = 24 - 15 = 9
So, W(4) = 9

For N=5:
5 right = 1
4 right = 0
3 right = 10
2 right = 20
1 right = 45 (not 50)
Total = 76
0 right = 44
W(5) = 44

etc.

Juice Part 2

Now let's look at the probability of getting all the guesses wrong.

If there are only two cans of juice then you either guess both correctly or both incorrectly. But with three cans, you could get one right and the other two wrong.

What is the probability of getting all three wrong

With 10 cans of juice, what is the probability that you will get none of them right

How about with 100 cans

Can you come up with a general formula for the probability of getting no matches at all when trying to guess the flavors of N cans of juice

Finally, what is the probability of getting just one right

we're sll going to have nightmares about soda cans tonight!

As in my previous post, let W(r) be the number of ways of getting r out of r wrong. This is the function Try is now asking for. Once it is found, the second part (getting just one right) is trivial.

There will be N cans that can be gotten right, and each will have W(N-1) ways of doing it. So, the final answer is:
N*W(N-1)/N! = W(N-1)/(N-1)!

We just need W(r).

Here's a recursive formula for W(r) (the number of ways of getting r out of r cans wrong):

W(0) = 1
w(1) = 0
For N>0,
W(N+1) = SUM(i=0 to N-1, (N!/i!)*W(i))

N W(N)
--------
1 0
2 1
3 2
4 9
5 44
6 265
7 1854
8 14833
9 133496
10 1334961

The probability of getting them all wrong is W(N)/N!.

N P(all N are wrong)
------------------------
1 0
2 .5
3 .33333
4 .375
5 .36667
6 .36806
7 .36786
8 .3678819
9 .3678792
10 .3678795

As N approaches infinity, P(all N wrong) approaches (rather quickly) 1/e = .367879441...

As noted in the previous post, P(one right) = W(N-1)/(N-1)! (for N>2). This is just the previous term in the sequence (the number of ways of getting all N-1 cans wrong). Therefore, it also approaches 1/e very quickly.

After a Nap.

I rode the bike (see avatar) down along the Tug Valley Today---didn't see any sign of the Hatfield/McCoy Feud. I did see Devil Ans' Hatfield 13 ½ foot Italian Marble Statue and Headstone. No sign of Randall McCoy's grave.

Series
[size=7]a) 1,4,9,16,25,36,……,n^2
b) 1,2,3,24,120,720,5040,……..,n![/size]

Flying over Pyrenees
[size=7]D=(Va+Vw)to
D=(Va-Vw)tb
2D=(Va+Vw)to+(Va-Vw)tb=Vato+Vatb+Vw(to-tb)
and
Vave=2D/(to+tb)= [Vato+Vatb+Vw(to-tb)]/(to+tb)
dVave/dVw=0=(to-tb)/(to+tb)
so Vave is max when to=tb and Vw=0
so if there is any wind speed correction (Vw) it will take longer.[/size]

Rings
[size=7]A+B=23
B+C=33
A+C=30
2A+2B+2C=86
A+B+C=43[/size]

3 digit Divisible number
[size=7]n=m*7*8
and n<1000
1000/(7*8)=17.86….
so m=17
n=17*7*8=952[/size]

Alpha Club
[size=7]240/(n-1)=240/n+1
so
n^2-n-240=0
(n-16)(n+15)=0
n=16 or n=-15
negative numbers do not make sanity check so
n=16
so the number of eligible members is n-1=15 (each gets $16) [/size]

Triple Sums of 10
[size=7]Including 0 as counting number
(0,1,9)(0,2,8)(0,3,7)(0.4.6)(0,5,5) 5
(1,1,8)(1,2,7)(1,3,5)(1,4,5) 4
(2,2,6)(2,3,5)(2,4,4,) 3
(3,4,3) 1
5+4+3+1=13
13 with 0 as counting #
8 if zero is not counting #[/size]

Number
[size=7]N/3=N-8
N=3N-24
2N=24
N=12[/size]

Still pondering the can problem. Us hillbillies have to take our boots off when ciphering numbers greater than 10.

Rap c∫;?/

Mark:

SERIES
a) N^2

This is the sequence of square numbers. It may be written like this:
12, 22, 32, 42, 52,
and it continues
12, 22, 32, 42, 52,62, 72, 82,
which is the same as
1, 4, 9, 16, 25, 36, 49, 64, ...


b) N!

These are factorials. A factorial of a number is the product of all the postitive integers from 1 to the given number. The factorial of 3, which is written 3!, is 1 * 2 * 3 = 6. 4! = 1 * 2 * 3 * 4 = 24. Our series of factorial may be written as
1!, 2!, 3!, 4 !, 5!, 6!, ...
and we may continue it:
1!, 2!, 3!, 4 !, 5!, 6!, 7!, 8!, 9! ...
which equals
1, 2, 6, 24, 120, 720, 5040, 40320, 362880,...
Factorials get very big very quickly.


FLYING
Longer

Many people immediately answer that the flying time is the same. (But not Mark) What you lose because of a headwind in one direction should be the same as what you gain due to a tailwind on the return trip. But if you do the calculations, you'll see that this isn't the case.

With no wind, the round trip will take 4 hours: 1600 kilometers at 800 kilometers per hour means the trip is two hours each way.

With a 200 kph headwind, the ground speed is 600 kph. The time it takes to cover 1600 km is
1600 km / 600 kph = 2.67 hours or 2 hours and 40 minutes
In the other direction the ground speed is 1000 kph so the time it takes to go 1600 km is
1600 km / 1000 kph = 1.67 h, or 1 hour and 40 minutes
The total travel time is 4 hours and 20 minutes.

Let's say that the wind speed is increased to 400 kph. Then the ground speed with a tailwind is 1200 kph, and the travel time in that direction is
1600 km / 1200 kph = 1.33 h or 1 hour and 20 minutes
In the other direction the ground speed is 400 kph so the travel time is
1600 km / 400 kph = 4 hours.
The total travel time is 5 hours and 20 minutes. This is an hour longer than with the wind at 200 kph.


TARGET
43

7 AND 8
952

ALPHA CLUB
15

TRIPLES
8

N
12


Rap:

Series
a) 1,4,9,16,25,36,……,n^2
b) 1,2,3,24,120,720,5040,……..,n!

Flying over Pyrenees
D=(Va+Vw)to
D=(Va-Vw)tb
2D=(Va+Vw)to+(Va-Vw)tb=Vato+Vatb+Vw(to-tb)
and
Vave=2D/(to+tb)= [Vato+Vatb+Vw(to-tb)]/(to+tb)
dVave/dVw=0=(to-tb)/(to+tb)
so Vave is max when to=tb and Vw=0
so if there is any wind speed correction (Vw) it will take longer.

Rings
A+B=23
B+C=33
A+C=30
2A+2B+2C=86
A+B+C=43

3 digit Divisible number
n=m*7*8
and n<1000
1000/(7*8)=17.86….
so m=17
n=17*7*8=952

Alpha Club
240/(n-1)=240/n+1
so
n^2-n-240=0
(n-16)(n+15)=0
n=16 or n=-15
negative numbers do not make sanity check so
n=16
so the number of eligible members is n-1=15 (each gets $16)

*Go with the $15*


Triple Sums of 10
Including 0 as counting number
(0,1,9)(0,2,8)(0,3,7)(0.4.6)(0,5,5) 5
(1,1,8)(1,2,7)(1,3,5)(1,4,5) 4
(2,2,6)(2,3,5)(2,4,4,) 3
(3,4,3) 1
5+4+3+1=13
13 with 0 as counting #
8 if zero is not counting #

Number
N/3=N-8
N=3N-24
2N=24
N=12

Rap wrote, "I rode the bike (see avatar) down along the Tug Valley Today---didn't see any sign of the Hatfield/McCoy Feud."

You lucky thang, they have good bass in the Tug river. Did you see Williamson's "Coal House"?
As fur them feudin, don't cross into Pike County. You have been warned!

"No sign of Randall McCoy's grave."

If memory serves, he dun survived the feud. Still I reckon iffin he deed, he would be deed by now.

"Us hillbillies have to take our boots off when ciphering numbers greater than 10."

I never did meet no redneck who could cipher nothing other than ?'moonshine'


Juice Part 2 Answer (sorta like Mark's)

If you have only two cans of juice, the probability of getting them both wrong is the same as the probability of getting them both right. This is because there are only two possible arrangements:

apple orange /orange apple

One is all right. The other is wrong.

With 3 flavors there are 6 possible arrangements:


Let's say that the first one is correct. Then there are 2 arrangements that are all wrong:

orange grape apple /grape orange apple

The other arrangements have one can in the correct place and the other two wrong. So the probability of guessing all wrong is 2 / 6 or 1 / 3.

In general, the question is: Given any sequence of n distinct objects, in how many ways can those objects be rearranged so that none of the objects is in its original position? This is known as the number of "derangements" of n, denoted by D(n). Another term for the number of derangements is the subfactorial denoted !n.

D(1)= !1 = 0 and D(2)=!2 = 1

One way (not the most efficient) of determining the value of D(n) for any given n is by means of the inclusion/exclusion principle. For example, with n=5 we begin with 5! possible re-arrangements, but of those we know that 4! leave the first object unmoved, and 4! leave the 2nd object unmoved, and so on. Thus there are 5*4! that leave one specific object unmoved, so we have to subtract this number. However, in doing so we have subtracted some of the arrangements twice, because some of them leave both the 1st AND the 2nd element unchanged (for example). So we need to add back the number of arrangements that leave any two of the objects unmoved, of which there are C(5,2)*3! (There's an explanation of what C means at the bottom of this page.) But then we need to subtract the number of arrangements that leave THREE objects unmoved, and so on. The final result is

D(5) = 1*5! - 5*4! + 10*3! - 10*2! + 5*1! - 1*0!
= 44

In general we have

n j / n \
D(n)= SUM (-1) ( ) (n-j)!
j=0 \ j /

If we examine the effect of multiplying each term of this sum by n+1, we see that the value of D(n) can be computed from the value of D(n-1) by the recurrence formula

D(n) = n D(n-1) + (-1)^n

The first several values of D(n) for n=1,2,3,... are

0, 1, 2, 9, 44, 265, 1854, 14833, ...

The Encyclopedia of Integer Sequences (Sloane & Plouffe) lists the values up to D(17), and also notes the exponential generating function

1 inf x^n
--------- = SUM D(n) ---
(1-x) e^x n=0 n!




C(5,2) means the number of possible combinations of two elements at a time selected from five elements.

………….. 5!
C(5,2) = ------
……..2! * 3!

……………. n!
C(n,k) = ------
………..k! * (n-k)!





A cube has six faces, each a square. So far so good.
How many different patterns of six squares are there that can be folded into a cube

I bet you never expected that question!

Version 1
A spider is climbing slowly up a wall. After one hour it is half way to the top. After one more hour it covers half the remaining distance so it is 3/4 of the way to the top. In the next hour it again covers half the remaining distance and is now 7/8 of the way to the top. If this pattern continues, how long will it take the spider to reach the top of the wall

Version 2
Now suppose that the spider again climbs half way to the top in the first hour. But it then covers half the remaining distance in half an hour and is 3/4 of the way to the top in 1 1/2 hours. Then it covers half of the remaining distance in 15 minutes. Again, continuing the pattern, how long will it take to reach the top


What is ½ + .1/2 + 1/.2 =

[size=7]CUBE
Excluding rotations, but including reflections, I have found 20.

SPIDER VERSION 1
It will never reach the top, but it can get as close as it wants to the top.

SPIDER VERSION 2
2 hours and an infinite number of movement intervals.

ADDITION
5.55[/size]

Cube Folding.
[size=7][size=7]I count eleven, which I subdivide into, Trivial (2),Easy (4), Variations on Easy(3), and Not So Obvious(2)

X's are cube sides, O's are Not there (except for format

1-Trivial)
OOOX
XXXX
OOOX
2-Trivial)
OOXO
XXXX
OOXO
3-Easy)
XOOO
XXXX
OXOO
4-Easy)
OOXO
XXXX
OXOO
5_Easy)
XOOO
XXXX
OOXO
6-Easy)
XOOO
XXXX
OOOX
7-Variations on easy)
OOXX
XXXO
OOXO
8-Variations on easy)
XXOO
OXXO
OOXX
9-variation on easy)
OOXX
XXXO
OXOO
10- Not so obvious)
XXXOO
OOXXX
11-Not so obvious)
OOOXX
XXXOO
XOOOO[/size]

Spider Version 1
[size=7]Never. The Spider's name is Zeno and he never reaches his limit[/size]

Spider Version 2
[size=7]Two hours[/size]

Sum
[size=7]½+1/20+5=5 11/20[/size]

BTW, Try, Gaulley season is fast approaching.

Rap c∫;?/

Mark:

CUBE
Excluding rotations, but including reflections, I have found 20.

Excluding r/r I only have 7.
Rap has 11.

I will have to find the shapes.

SPIDER VERSION 1
It will never reach the top, but it can get as close as it wants to the top.

SPIDER VERSION 2
2 hours and an infinite number of movement intervals.

ADDITION
5.55

The Speed of Spiders
The spider will never reach the top of the wall. Each step of the way it covers half the remaining distance to the top so there is always some distance to go. In the next step only half of that distance is traveled, still leaving a bit more to cover. This is true for both versions of the puzzle.
But, there is a concept in mathematics of a "limit." This is a number or point that is approached, even if it is never reached. We can say that as the number of steps the spider travels approaches infinity, the distance traveled approaches the height of the wall as a limit. If the spider took an infinite number of steps it would reach the top.

Now with that perspective there is a difference between the two versions. If each step takes an hour, then it would take an infinite number of steps to reach the top in an infinite amount of time. But in Version 2 each step takes half as long as the previous step. The total time to reach the top approaches two hours as a limit.


Rap:

Cube Folding.
I count eleven, which I subdivide into, Trivial (2),Easy (4), Variations on Easy(3), and Not So Obvious(2)


Spider Version 1
Never. The Spider's name is Zeno and he never reaches his limit

Spider Version 2
Two hours

Sum
½+1/20+5=5 11/20

BTW, Try, Gaulley season is fast approaching.

?'Oh, you mean the "Beast of the East", Starts September 9th I think. Have you done both the upper and the lower sections?'



The average of five numbers is 6. If one of the five numbers is removed, the average of the four remaining numbers is 7. What is the value of the number that was removed


The serial number of my camera is a four-digit number less than 5,000 and contains the digits 2, 3, 5, and 8 but not necessarily in that order. The "3" is next to the "8", the "2" is not next to the "3", and the "5" is not next to the "2". What is the serial number


When asked how many bananas he had, the collector said: If I arrange them in stacks of five, none are left over. If I arrange them in stacks of six, none are left over. If I arrange them in stacks of seven, one is left over.

What is the least number of bananas he could have

No camels were hurt in the formation of this riddle.


When a natural number is multiplied by itself, the result is a perfect square. For example 1, 4, and 9 are perfect squares because 1 x 1 = 1, 2x2 = 4, and 3x3 = 9.

How many perfect squares are less than 10,000

Both sections. But prefer the upper, it's more intense. Two years ago the river was closed on opening day (too much water). I went the second day and got to swim the strainer. The boat flipped at iron ring. Gawd! What a good day to live!

Tried the Upper New in a canoe last summer. Spent some time swimming there too. I've never kayaked, guess that will be on my gravestone, but my partner and I have done an eskimo roll in a canoe (wasn't intentional).

Average
[size=7]n=5*6-4*7=28[/size]

Camera
[size=7]Serial Number 2835[/size]

Bananas
[size=7]n=m*5*6=m*30
30mod7=2 group 2,4,6,1,3,5,0 m=4
m*20=130
120 bananas[/size]

Perfect Square
[size=7]n=10,100=100^2
so for n<100 there's 99 unless 0^2 is included, then it's 100[/size]

Rap c∫;?/

CUBE
Rap and I have the same answers. Nine of his can be reflected to yield new patterns.
#11 should be:
OOXX
XXXO
XOOO

[size=7]FIVE NUMBERS
2

SERIAL NUMBER
2835

BANANAS
120

SQUARES
99[/size]

Cubes - Six sides - Seven ways - No r/r. Otherwise 28 ways.





Rap wrote, "I've never kayaked, guess that will be on my gravestone, but my partner and I have done an eskimo roll in a canoe (wasn't intentional)."


That reminded me of a birthday trip on a cold October day and an Eskimo who tried to keep me warm using a parafin stove, whilst cutting the cake.
The resulting fire lead to the punch line - 'You can't have your kayak and heat it too'.