The worlds first riddle!

How can Mark’s answer be right?

Any move decreases the number of remaining sections by 1. Therefore, after k moves, the total number of remaining sections will be 500 − k irrespective of an order in which the sections are assembled.

Hence 499 moves will be made before the entire puzzle is assembled; so simple.



Rap, you crack me up; I was looking up the Law of reciprocity which in number theory, the law of quadratic reciprocity is a theorem about modular arithmetic which gives conditions for the solvability of quadratic equations modulo prime numbers….

And do you know what! I’m not even in my prime!
Besides that, I thought you had misspelled Ciao!



A Fake among Eight Coins (Re-visited)

There are eight identical-looking coins; one of these coins is counterfeit and is known to be lighter than the genuine coins.

What is the minimum number of weighings needed to identify the fake coin with a two-pan balance scale without weights?


Something for the weekend!

I have found the A2K book of famous sayings, original quotations and witty rejoinder’s; needless to say the pages are blank!

However the pages of this well bound book are numbered sequentially starting with 1.

If the total number of decimal digits used is equal to 1578, how many pages are there in this priceless tome?


Have a good weekend guys; I’m planning to spend a few days under the freeway off ramp where I grewed up.

This can be done in 2 weighings. Weigh any three coins against any other three. If one side is lighter, the coin lies among those three. If neither side is lighter, the coin is one of the two you didn't put on the balance.

If you have two coins left that could be counterfeit, simply weigh them against each other to see which is fake.

If you have three possible fakes, weigh two against each other. If one is lighter it is counterfeit. If neither is lighter, the third is counterfeit.

coins: you could have had 9 coins and still used only 2 weighings
digits: the first 99 pages require 189 digits - that leaves 1389/3=463 more pages - total number of pages is 562

Kolyo, it is said that many seek enlightenment by answering a question posed here; in your case you have found Nirvana – your answer is perfect.
Welcome and please be seated at the top table.

Mark, you are correct regarding the coins.

“Digits: the first 99 pages require 189 digits - that leaves 1389/3=463 more pages - total number of pages is 562”

And also with the total number of pages; and your answer is much more succinct than mine.


Let D(n) be the total number of decimal digits in the first n positive integers
(book pages). The first nine numbers are one-digit, therefore D(n) = n for 1 ≤
n ≤ 9. The next 90 numbers from 10 to 99, inclusive, are two-digits. Hence,
D(n) = 9 + 2(n − 9) for 10 ≤ n ≤ 99.
The maximal value of D(n) for this range is D(99) = 189, which means that some
three-digit numbers are needed to reach the total digit count of 1578 given in the puzzle. There are 900 three-digit decimals, which leads to the formula
D(n) = 189 + 3(n − 99) for 100 ≤ n ≤ 999.
To answer the puzzle’s question, we need to solve the equation
189 + 3(n − 99) = 1578.

Its solution is n = 562. Ha-ha, just follow Mark’s example!


If we generate a list of all “words” made of letters G, I, N, R, T, and U in lexicographic order starting with GINRTU and ending with UTRNIG……

What position in the list will be occupied by TURING?



More damn coins!

The object of this conundrum is to place the LARGEST possible number of coins at points of the eight-pointed star depicted below.



The coins must be placed one after another, with the following restrictions:

(i) A coin needs to be placed first on any unoccupied point and then moved along a straight line to another unoccupied point, and…
(ii) Once a coin has been positioned in this manner, it cannot be moved again; I mean, totally verboten!

For example, we can start by placing the first coin on point 6 and then moving it to point 1 (denoted 6→1), where the coin will have to remain. Kapish!

Please list the order of your moves, the first to place all eight will receive the highest award A2K can offer – Thank you.

Oh, I will run nekkid round Times Square carrying a chicken.

Turing:
There are 4*5!=480 words before the first word that starts with T.
There are another 4*4!=96 words before the first word that starts with TU.
There are another 3*3!=18 words before the first word that starts with TUR.
There are another 1*2!=2 words before the first word that starts with TURI
There are another 1*1! words before the first word that starts with TURIN
Turing is the first and only word that starts with TURIN.
Therefore, there are 597 words before Turing.
Therefore, Turing is the 598th word.

Another approach:
There are 5*5!=600 words before the first word that starts with U.
There are only 2 words after Turing before the first word that starts with U:
599: TURNGI
600: TURNIG
Therefore, Turing is the 598th word.

Coins: The second part of (i) can't be accomplished with the 8th coin. Assuming that's OK:
6->1
3->6
8->3
5->8
2->5
7->2
4->7
4 (stays at 4)

Yet another approach:

The total number of words made up of the six letters is equal to 6! = 6 • 5 •4 • 3 • 2 • 1 = 720.

The “words” following TURING in the alphabetically ordered list are either of the form U ∗ ∗ ∗ ∗ ∗ or of the form TURN∗∗, where a ∗ stands for one of the six letters not already in the “word.” Since these letters can be in any order, there are 5! = 5 • 4 • 3 • 2 • 1 and 2! = 2 • 1 of them, respectively.
Thus, the total number of “words” following TURING in the alphabetically ordered list is equal to 5! + 2! = 120 + 2 = 122.

This means that TURING will be in position 720 − 122 = 598 in the list, if its elements are numbered from 1 to 720.







You are correct…. And the eighth coin cannot be placed due to the restrictions. Thank goodness the nation is spared the sight of me and my chicken…Ugh! What a fowl thought.




As y’all may know; I’m maybe planning to be a big-time rock star and have just returned from busking down in the subway and have amassed a fine collection of 33 coins.

These consist of nickels, dimes, and quarters, the total value of which is $3.30. This is not bad for a morning’s work; so up yours IRS!

If there are three times as many nickels as quarters, and one-half as many dimes as nickels, how many coins of each kind are there?



On checking my wallet after the last gig… it contains the same number of pennies, nickels, and dimes. The coins total $1.44.

How many of each type of coin does my wallet contain?


Carnegie Hall here I come….. God willin' and the creek don't rise!

Six quarters
twenty one nickels
twenty dimes
nineteen pennys

Rap

33 coins:
Since for each 6 nickels there are 2 quarters and 3 dimes (a total of 11 coins) and we need 33 coins, there must be 3*6=18 nickels, 3*2=6 quarters, and 3*3=9 dimes.

$1.44
Since a penny, a nickel, and a dime are 16 cents, there must be 144/16=9 of each.

Agree with the 18 nickles, 9 dimes and 6 quarters for 33 & $3.30,
but 33 & $1.44 has another solution (19 pennys, 11 dimes & 3 nickels)

Rap

On checking my wallet after the last gig… it contains the same number of pennies, nickels, and dimes.

oops!

Rqp

Mark nails both questions (as one would expect), and in a most enlightened way; but can his run last!

Cheer up Rap; you were right about the six quarters. It is not about the winning, but the taking part.

Show me a person who has never made a mistake, and I’ll show you someone who has never made anything! Just blame Tricky Dicky for the wording.


I'll start by picking and defining a variable, and then I'll use translation to convert this exercise into mathematical expressions.
Nickels are defined in terms of quarters, and dimes are defined in terms of nickels, so I'll pick a variable to stand for the number of quarters, and then work from there:
number of quarters: q
number of nickels: 3q
number of dimes: (½)(3q) = (3/2)q
There is a total of 33 coins, so:
q + 3q + (3/2)q = 33
4q + (3/2)q = 33
8q + 3q = 66
11q = 66
q = 6

Then there are six quarters, and I can work backwards to figure out that there are 9 dimes and 18 nickels.



“I have redde, I know not where, these verses. A woman, an asse, and a walnut tree, Bring the more fruit the more beaten they bee.”

[1581 G. Pettie tr. S. Guazzo's Civil Conversation iii. 20]

I totally repudiate this Republican fillossofee; so I have decided to recycle my old whacking stick in an eco-friendly way…

My stick is 100 units long and needs to be cut into 100 unit pieces.

What is the minimum number of cuts required if you are allowed to cut several stick pieces at the same time?

ceiling(log2(100)) = 7

Mark kicks butt:
ceiling(log2(100)) = 7

And in English…
The minimum number of cuts for a 100-unit stick is seven.

Since cutting several pieces of a given stick at the same time is allowed, we need to concern ourselves only with finding a cutting algorithm that reduces the size of the longest piece present to size 1.

This implies that on each iteration an optimal algorithm must cut the longest piece—and simultaneously all the other pieces whose size is greater than 1—by half (or as close to this as possible).

That is, if the length l of a piece is even, it is cut into two pieces of length l/2; if l is odd and greater than 1, it is cut into pieces of lengths; l/2= (l + 1)/2 and l/2= (l − 1)/2, respectively.

The iterations stop after the longest—and, hence, all the other pieces of the stick has length 1.


I think his answer is worth a drink, if…..

Given an 8-pint jug full of beer and two empty jugs of 5 and 3 pint capacity; is it possible to get exactly 4 pints of beer in one of the jugs by completely filling up and/or emptying jugs into others?

If after suppin’ a few beers while contemplating this conundrum one came to the hazy conclusion that it may be possible…..

How many operations would be necessary to achieve the objective?


Should any A2K Temperance Movement members find the puzzle abhorrent, please fill the jug with Kool-Aid; and if you decide that 8, 5 and 3 can never equal four – I will be delighted.

beer: 6 operations
8 5 3
------
8 0 0 (starting point)
3 5 0 (pour 5 from 8 to 5)
3 2 3 (pour 3 from 5 to 3)
6 2 0 (pour 3 from 3 to 8)
6 0 2 (pour 2 from 5 to 3)
1 5 2 (pour 5 from 8 to 5)
1 4 3 (pour 1 from 5 to 3)

I can add nothing to Mark’s excellent answer except: Rap drank the puzzle and did so in one operation!

Now, before you say, ‘That is Firkin amazing’, may I point out that…..

One Fluid Ounce (Fl. oz) is 1/160 of a Gallon
One Gill is 5 Fl. oz or 1/32 of a Gallon
One Pint is 4 Gills or 20 Fl. oz or 1/8 of a Gallon
One Quart is 2 Pints or 8 Gills or 40 Fl. oz or 1/4 of a Gallon
One Firkin is 9 Gallons
One Kilderkin is 2 Firkins or 18 Gallons
One Barrel is 2 Kilderkins or 36 Gallons

However to be as smart as Mark you have to remember that…

An Imperial British Gallon was defined as the volume occupied by exactly 10 pounds of water of density 0.988859 gramme per millilitre weighed in air of density 0.001217 gramme per millilitre against weights of density 8.136 grammes per millilitre - got that?

And for all those decadent decimalised folk out there: One Pint is 0.577 644 Litres


BTW are shrinks physic? I only ask because I walked into my shrink’s office wrapped in clear cellophane (long story) and asked for an urgent appointment and she said, “Obviously; I can see your nuts."

Ask any ‘expert’ and they will tell you this cannot be done; if you ask me, I would say, Meh!

(a) You have 20 black nuts and 16 white nuts in a bag. You repeat the following operation until a single nut is left in the bag.

You remove two nuts at a time. If they are of the same color, you add a black nut to the bag; if they are of different colors, you add a white nut to the bag.

Can you predict the color of the last nut left in the bag?

(b) Answer the same question if there are 20 black nuts and 15 white nuts to start with.

Have a good weekend guys.

(a) black
(b) white

Because...every move reduces the number of white nuts by either 0 or 2.
Remove two black nuts, and add back a black nut; the net change to white nuts is zero.
Remove two white nuts, and add back a black nut; the net change to white nuts is two.
Remove one of each, and add back a white nut; the net change to white nuts is zero, again.

nuts: The operations preserve the parity of white and alter the parity of black. In the first case, since white starts even, it must remain even. Therefore, black will be left. In the second case, since white starts odd, it must remain odd. Therefore, white will be left.

I may have seen better answers, but I don’t remember when. Delight in their eloquence.


Kolyo
(a) black
(b) white

Because...every move reduces the number of white nuts by either 0 or 2.
Remove two black nuts, and add back a black nut; the net change to white nuts is zero.

Mark
nuts:
The operations preserve the parity of white and alter the parity of black. In the first case, since white starts even, it must remain even. Therefore, black will be left. In the second case, since white starts odd, it must remain odd. Therefore, white will be left.




An A2K moderator put a pair of hamsters in a place surrounded on all sides by a wall.

The initial pair of hamsters (male and female) are new born.
All hamsters’ pairs are not fertile during their first month of life but give birth to one new male/female pair at the end of the second month and every month thereafter.

How many pairs of hamsters will be there in a year?


I have been informed that to comply with inclusivity requirements where sexual procreation is determined by selective breeding, I must include a question for alternate life choices.

Therefore not wishing to offend any Democrats…


There are two queer hamsters new born… Yackety yak…. Yackety yak…
How many pairs of hamsters will be there in a year?


Please note: No hamsters were hurt in the composition of this puzzle.

hetero hamsters: Per Leonardo of Pisa, 233 pairs at the end of the 12th month.
homo hamsters: Only 1 pair, but their enclosure is decorated very fashionably.