The worlds first riddle!

Mark:

KNUTH
[sqrt([sqrt(sqrt((3!)!))]!)]

3! = 6
6! = 720
sqrt(720) = 26.83
sqrt(26.83) = 5.18
[5.18] = 5
5! = 120
sqrt(120) = 10.95
[10.95] = 10


As you may have noted, I half gave the answer. Since [sqrt(26)] = 5 and [sqrt(5)!] = [sqrt(120)] = 10, we get one solution just by stringing everything together.


Here's one from the USA Mathematical Talent Search; it's not easy, but it's certainly doable.

For a positive integer n, let P(n) be the product of the nonzero base 10 digits of n. Call n "prodigitious" if P(n) divides n.

What is the maximum number of consecutive prodigititious positive integers n


Hint: [size=7]The answer is not 12[/size]

[size=8]PRODIGITIOUS
Previously solved after receiving a hint:

http://www.able2know.com/forums/a2k-post1056926.html#1056926[/size]

Ok, I have junked that file as it seems to have some old stuff. Try this:


The three of us made some bets:

First, Waldo won from Molly as much as Waldo had originally.
Next, Molly won from Spike as much as Molly then had left.
Finally, Spike won from Waldo as much as Spike then had left.
We ended up having equal amounts of money.
I started with 50 cents.

Who am I

you're molly.... waldo starts with 3x money, molly with 5x, and spike with 4x. the only way for there to not be fractional pennies would be if waldo has 30 cents, molly 50, and spike 40.

Thoh:

you're molly.... waldo starts with 3x money, molly with 5x, and spike with 4x. the only way for there to not be fractional pennies would be if waldo has 30 cents, molly 50, and spike 40.


Don't make it look so easy dude, mix it up like this:

Let Waldo start with w, Molly with m, and Spike with s. Following the above account we get the following progression of monies:
• W = 2*w, M = m-w
• M = 2*(m-w), S = s-(m-w) = s+w-m
• S = 2*(s+w-m), W = 2*w-(s+w-m) = w+m-s

And so Waldo finished with w+m-s, Molly with 2*(m-w), and Spike with 2*(s+w-m). Since these must all be equal, we have three equations and three unknowns, so solve; this gives us that 4*m = 5*s and 3*s = 4*w. Now, if s = 1/2 this implies that w = 3/8 = 37.5 cents, an impossible amount of money to start with. If w = 1/2 this implies that s = 2/3, again an impossible amount to start with. Finally, if m = 1/2 this implies that s = 2/5 = 40 cents, w = 3/10 = 30 cents, which works. It follows that I am Molly.



To reward Mark for his contributions over the years, A2k wishes to build a great pyramid for his internment. The structure will have a square base and be solidly composed of cubical stone blocks. Each level of the pyramid contains one less block per side as the pyramid rises.

Thoh (project manager) has available an initial work force of 35,000 A2K workers. Each morning (at 6am) the available labor pool is divided into work crews of 17 workers each. Any remainder that cannot form a full crew gets the day off but are available the following day.

Each crew can lay one block of the pyramid each day. Unfortunately, the hard work causes the death of one member of each crew each day. (Don't worry, we are insured) Work ceases on the project when it can be determined that there will be insufficient a2k'ers available to raise the pyramid one more level. Each stone block measures 3 meters per side.

How long (in days) will it take to construct Mark's pyramid

How tall will it be

How many of the original workers survive the construction


Don't ask your Mommy for help

[size=8]PYRAMID
It will take 53 days to build a pyramid 46 levels (138m) high. 1489 crew members survived.[/size]

Mark:

PYRAMID

It will take 53 days to build a pyramid 46 levels
(138m) high. 1489 crew members survived.


This classic was printed in the October 1992 issue of PuzzleSIGns, the Mensa puzzle SIG publication. I don't know the original source.

Each block added results in the death of exactly one slave, and when there are less than 17 slaves no more blocks may be added, so the pyramid consists of at most 35000-16 = 34984 blocks.

The first level needs 1 block, the next 4, and in general the ith level requires i^2 blocks, and so a pyramid n levels tall requires 1 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 blocks. This is between n^3/3 and (n+1)^3/3, and solving n^3/3 = 34984 we get that the maximum number of levels is either 46 or 47, with a quick check showing that 46 is correct. It follows that the pyramid is 46*3 = 138 meters tall, consists of 33511 blocks, and the number of crew members surviving the construction is 35000-33511=1489.

To compute the number of days required for construction, note that if there are m slaves on a given day the number of blocks added is between m/17 and m/17 - 16/17, and given k days it can be easily shown that the number of blocks added is between m/17 + (16/17)m/17 + ... + (16/17)^{k-1} m/17 and m/17 + (16/17)m/17 + ... + (16/17)^{k-1} m/17 - k*16/17; summing the geometric series we get m*(1-(16/17)^k), and so to find the number of days required we solve 35000*(1-(16/17)^k) = 33511 and get that (16/17)^k = .04254 or k=52.08.

Since this came from an upper bound on the number of blocks, it's a lower bound on the number of days; thus, it'll take at least 53 days to build the pyramid. Checking the lower bound, we see that the number of blocks that can be added in 53 days is greater than 33542, and so the pyramid can indeed be built in 53 days. Although I hope it will not be required for at least the next 53 years.



Try is a bit absentminded, and the only way he can remember his own phone number is that if you divide it by its reverse, you get an integer greater than one.

What's Try's phone number

[size=8]TRY'S PHONE NUMBER
Try can be reached at 879-9912 or 989-9901. He probably has two residences with separate wives that don't know about each other...[/size]

Mark:

TRY'S PHONE NUMBER

Try can be reached at 879-9912 or 989-9901.

"He probably has two residences with separate wives that don't know about each other..."

Thanks buddy, the phones been ringing off the hook, pity you forgot 1-800-wifeletts. Lucky you missed all my non listed numbers. :wink:




Mark, Thoh, Try, two wedges of cheddar, two wedges of mozzarella and Rufus the dog are going to go to the annual Hicksville cheese competition and dog show.

Mark's little sports car can only seat two objects at a time; Mark, Thoh and Try can all drive the car.

If Mark is not around, Thoh will eat the cheddar; if Thoh is not around, Mark will eat the mozzarella. If Try is not around, Rufus will eat the cheese and bite Mark and Thoh.

Can they all get to Hicksville without anything bad happening (Y/N)

If so, what's the smallest number of trips needed
(each way counts as one trip)


If you need assistance ring 1-800-MARK (Local call rates apply)

CHEESEDOG

17 trips. In the chart below, the dashes in the "Car" column indicate the direction. Greater than and less than didn't work.

Mark:

CHEESEDOG

17 trips

(See his answer for how it's done)

His answer is perfect, no doubt he wrote a little programme to sort it all out. If your business is looking for answers, seek him out.

The other method is to do this in reverse. Another way of looking at it is that we're swapping cheddar and mozzarella, Thoh and Mark, since from the point of view of the other restrictions this changes nothing.


Try is having a party and has 50 guests, among whom is his brother Cleetus.
Cleetus starts a rumor about Try; a person hearing this rumor for the first time will then tell another person chosen uniformly at random the rumor, with the exceptions that no one will tell the rumor to Try or to the person they heard it from. If a person who already knows the rumor hears it again, they will not tell it again.

What's the probability that everyone, except Try, will hear the rumor before it stops propagating

I did NOT write a program. However, about half way through, I did think that it would be interesting to write a program to solve this sort of problem.

[size=8]RUMOR
Might as well be zero. I get 2.4793E-20.[/size]

This puzzle is brought to you by our sponsors; French cheese eating S.M's.

[size=8]CHEESE

Mark: 15 + 18 = 33
Cleetus: 16 + 19 + 31 = 66

mozzarella = 20[/size]

Mark:

CHEESE

Mark: 15 + 18 = 33
Cleetus: 16 + 19 + 31 = 66

mozzarella = 20


Mark must have bought at least 31 pounds of cheese, and so Cleetus must have bought three wedges since the two heaviest wedges together only total 51 pounds. Since Cleetus bought twice what Mark bought, the total amount of cheese bought by the two together equals 3 times that purchased by Mark alone.

Now, the total weight of all the wedges at the grocery divided by 3 leaves a remainder of 2, so the mozzarella wedge must leave a remainder of 2 when divided by 3. But the only such wedge is the 20 pound wedge, thus the mozzarella wedge must have weighed 20 pounds.



Mark and Cleetus were recently invited to a party attended by four other pairs of siblings, for a total of ten people. During the party various handshakes took place, but no person shook their own hand or the hand of their sibling.

At the end of the party Mark asked each person, including Cleetus, how many different people they shook hands with, and was surprised to note that every number was different!

How many hands did Cleetus shake

[size=8]HANDSHAKES
4

The nine different responses were 0-8. They are paired up (siblings) like this:
8-0
7-1
6-2
5-3
4-

The other 4 is Mark's number. Therefore, Cleetus was the reported 4.[/size]

Mark:

HANDSHAKES
4

The nine different responses were 0-8. They are paired up (siblings) like this:
8-0
7-1
6-2
5-3
4-

The other 4 is Mark's number. Therefore, Cleetus was the reported 4.



Yup, you have done it again;

Since each person shook a different number of hands, the totals must range from 0 to 8. Now the person who shook 8 hands must have shaken the hand of everyone except their sibling, so their sibling must be the one who shook 0 hands. It follows that the persons who shook 1 and 7 hands must be siblings, since we must have that 1 only shook with 8, 7 with everyone but 0 and 1.

In a similar fashion the persons who shook 2 and 6 hands must be siblings and the persons who shook 3 and 5 hands must be siblings. Since the only number left is 4, it follows that Cleetus must have shaken 4 hands.




As most of you know, once every year or so I like to post a riddle that is so fiendishly difficult it will defy even the best. This is one of those…


At a school camp, five schools; Aldhouse, Bedminster, Chartry, Radford and Rugenham were represented. The smallest contingent from the five schools was greater than 20 but less than 30.

Aldhouse sent two less than half of the Rugenham contingent. The Radford and Rugenham contingents together were 14 greater than the combined Bedminster and Chartry contingents.

The Bedminster and Rugenham contingents together were two short of half the total complement from the five schools while the Chartry and Radford contingents combined mustered 13/32 of that total.

What was the strength of each contingent

[size=8]FIENDISHLY DIFFICULT
Aldhouse = 26
Bedminster = 72
Chartry = 36
Radford = 68
Rugenham = 54[/size]

Mark:

FIENDISHLY DIFFICULT

Aldhouse = 26
Bedminster = 72
Chartry = 36
Radford = 68
Rugenham = 54




I think both calculations come up with the answer


Aldhouse 26
Bedminster 70
Chartry 38
Radford 66
Rugenham 56



However, if you are unsure, run this little beauty:

model 'Fiendishly'


uses ' mmxprs '

declarations
x: array(1..5) of mpvar
m: mpvar
end-declarations

any:= x(1)

forall(i in 1..5)
fmin(i):= x(i) >= m
scona:= m >= 21
sconb:= m <= 29
cona:= x(1)=.5*x(5)-2
conb:= x(4)+x(5) = x(2)+x(3)+14
conc:= x(2)+x(5) = sum(i in 1..5).5*x(i)-2
cond:= x(3)+x(4) = sum(i in 1..5)(13/32)*x(i)

forall(i in 1..5)
x(i) is_integer

minimise(any)

forall(i in 1..5)
write(getsol(x(i)),' ')

end-model




Mark is a biology student. His project for this semester is measuring the effect of an increase in vitamin C in the diet of nine laboratory Coypu. Each Coypu will have a different diet supplement of 1 to 20 units. Fractions of a unit are not possible.

To get the maximum value for his experiment, Mark has decided that for any group of three Coypu, the supplements should not be in arithmetic progression. In other words, for three Coypu chosen at random, the biggest supplement less the middle supplement should be different from the middle supplement less the smallest supplement.

Thus, if two of the supplements were 7 and 13 units, no Coypu could have a supplement of 1, 10 or 19 units.

Can you help the dude out and find a set of supplements that Mark could use


Coypu supplied by: Gustav Farms Inc.

[size=8]COYPU
1, 2, 6, 7, 9, 14, 15, 18, 20

I did write a program for this one.[/size]