bonanova

This college student and his gf enter together and are taken to a cozy place for two where they're seated. An attractive woman walks over and asks what he's drinking. He orders a Corona. Noting his youthful appearance, the woman asks to see his license. He obliges, she nods her approval, and returns with his beer. After a fun conversation with his gf, the guy again feels a thirst coming on. He looks around and signals for service. A different woman comes by, this one not so attractive. Again he orders a Corona, and again he is carded. This time he is refused service. This is another adapted chestnut, so some may have heard it before. Spoilers, please.

Summary: I wonder whether it's more fun to make these puzzles or to solve them... First the close but no cigar awards ... Preflop - yeah, but you forgot that Q + D + E = 100 KlueMaster - you forgot to read the OP's invisible ink section [my bad ] that said Q, D and E are non-zero. Joey D - you get the most likely Google prowess award. And now the envelopes, please araver [repeating] the [claimed by OP to be unique] even answer. araver the [actually] unique odd answer HoustonHokie the Unadvertised Special - the second even answer. The puzzle is adapted from a published one that asked the question in a simpler way. I tried to toughen it up a bit. It gave the value of the Eagle [it wasn't about coins, but values were given: $0.25, $1 and $15] and asked how many of each would both total and cost 100. That source might be known to Joey D, because he came up with the resulting 56 41 3 solution. Which explains his award. OK why did I claim the even solution was unique? Because unconstrained there are a zillion solutions. My first try at limiting them was to specify the value of the Eagle to be less than $10. But that required further constraints. Then I spotted the neat fact that [Q-D]/E = even integer eliminated all the cases except two - one made V odd, the other made V even. Voila! I went with it. The flaw was I forgot about the V>10 answer I'd thrown into the wastebasket. Nice work all, and I exclude myself from that for the mistakes.

Bravo! Care to find the unique odd-dollar value?

This post was subsequently appended to the [qv] So I'm locking this thread.

Yes, this answer and variants have been given. I agree that this is a good "engineering" solution, being an engineer myself, I must agree that it works, and in the real world that's really all that matters. The OP leaves us with only this restriction on the solution, and it's ambiguous enough, IMHO to admit this one. namely, He will be only allowed to answer "BLACK" or "RED". If he says anything else you will ALL be executed immediately. It's clear to some of the problem solvers, but not to all, that the intent is to give a binary response, not embellished by differences in tone, loudness, time delay, voice pitch, or ... well or anything else. To other solvers, these embellishments are permitted, and the response is no longer binary. So there are two classes of solutions. Yours belongs to the latter. Welcome to the Den!

Although OP did not say this explicitly, the "certain numbers" of the three types of coins purchased are non-zero.

Ah, I understand. OK. The OP states "the purchase comprised a certain number of quarters and dollars ... and ... Eagles" So the question of interpretation is what purchase [used here as a noun] refers to. onelook.com gives us this: Quick definitions from WordNet (purchase) ▸ noun: the acquisition of something for payment ("They closed the purchase with a handshake") ▸ noun: a means of exerting influence or gaining advantage ("He could get no purchase on the situation") ▸ noun: something acquired by purchase It wouldn't be the first meaning, or the second. That leaves the third. That is, the Quarters, Dollars and Eagles are coins that were acquired by purchase. In none of the cases does purchase refer to something used as payment. In the puzzle, the payment was the $100 bill. So that's the intended sense of the puzzle - that certain numbers [non-zero numbers should be inferred] of Quarters, Dollars and Eagles, coins all, were purchased using a $100 bill [which was exact payment.] With the added information in my bad for leaving it out of OP, namely that the number of Quarters purchased equals the number of Dollars plus an even multiple of the number of Eagles and that the Eagle is worth a whole number, say n, of dollars, then there is a unique value for the Eagle coin [1] when n is odd and another unique value [2] when n is even. Hope that clarifies.

Yekangi is correct. The wording of the OP " ... I'll test all four of them for acidity. When I give you those results, you must then tell me ... " precludes some of the answers given.

methinks, krokce, k-man, phillip182 and MissKittten have it. methinks is first solver. MissKitten's observation is correct, this puzzle is a "chestnut" and has appeared in various forms previously. It retains a certain charm in that the wrong answer is seemingly so obvious.

Yes, there is more information. . Eagles are worth an even number of whole dollars.The number of Quarters equals the number of Dollars plus an even multiple of the number of Eagles.. With additional information comes a Bonus question: What would be the value of an Eagle if it were an ODD number of whole dollars, the other information remaining unchanged?

Nope. Perhaps deceptively simple?

It's finals day at the chem lab. Your instructor has placed on your table 13 numbered beakers containing colorless and odorless liquids. "The contents of one and only one beaker are slightly acidic," he tells you, "the others are neutral; they have a pH of exactly 7." He also hands you four clean flasks, labeled A, B, C and D. Put anything you like from the beakers into these flasks, and I'll test all four of them for acidity. When I give you those results, you must then tell me which of the 13 beakers is acidic. You never got an easier A.

As I recall, the man paid using a $100 bill.

A while back it was noted that a famous soft drink was 99% water. The other contents, that gave the drink its exquisite taste, weren't identified. One day a vat containing 1000 pounds of the drink was left uncovered, and due to evaporation its water content decreased to 98%. What was its decrease in weight?

Guy walks into a coin shop and buys 100 coins and pays $100 for them. The purchase comprised a certain number of quarters and dollars at face value, $0.25 and $1.00, respectively, and a certain number of gold coins known as Eagles. What's the unit price of an Eagle?

Imagine, or draw, or grab some play dough and make, a cube, 6" on a side. Select one corner and trace its three edges back to their nearest vertex. Slice off the corner of the cube, using the plane defined by these vertices. Do the same with three other corners, none of which is adjacent to a previously removed corner. What remains is a regular triangular pyramid, a tetrahedron. Its vertices are the four un-cut corners of the cube. What is the volume of this tetrahedron? Slice the tetrahedron in such a way that the newly created face is a square. What is the area of this square?

I'm not a whiz at Physics yet, so if I'm missing an important point... well, let me know. It depends. A black object will radiate to and absorb from its surroundings, eventually reaching the same temperature. That is also true of a red object. The difference is that a black object will do it more quickly. The slowest to equilibrate with their surroundings are white or reflective objects. That's why igloos sustain elevated interior temperatures, and shiny coffee pots keep their contents warm. If the prisoners' ambience is warmer than body temperature, a black hat will more quickly feel warm than will a red hat. Similarly it will more quickly feel cool than will a red hat, if the ambient temperature is below body temperature. But for this to be a life-saving color indicator, the head of each prisoner would have to have excellent temperature acuity, and each prisoner would have to know how fast a red hat would equilibrate, in order to know whether his [black?] hat had equilibrated more quickly. But even this would not suffice if the ambient temperature were the same as the prisoners' body temperature, or close to it. Then there would be no warming or cooling to perceive. And we haven't even touched on other mitigating factors related to temperature such as varying amounts of perspiration, a likely prospect for persons facing execution, if, as it evaporated, it cooled a prisoner's head more efficiently than a black hat might warm it. By comparison, keeping track of running parity seems simple and reliable. Isn't it the method you would choose?

I think this gives 19 people a definite survival and one person an 80-90% chance of survival.

If the incremental numbers are of a single digit the strategy simplifies greatly. As does finding the solution. Good job.

Hi 1up, and welcome to the Den. How many prisoners will be saved in your scenario? Is it a better result than 19?

In a we parsed a deck of 52 cards into sequentially higher and lower piles, and pondered the expected high card value in the low pile. The answer was found to be Let's say this answer is correct. Observe that the ratio r52 = 43.55 / 52 = 0.8375. Let's also assume that as the number of cards n in the deck increases without limit, the ratio rn converges to some value ro. I'm guessing there is a closed form expression for ro that involves the number e. Anyone care to derive it?

Here's a simple game between two players. Player A picks an integer p of at least four digits, like 1739. Player B picks an integer n of no more than two digits, like 23. Play proceeds as follows. Player A says a number from 1 to n, inclusive, like 16. Player B adds to it a number from 1 to n, inclusive, like 9, and says 25. Players alternate turns, adding to the running total numbers from 1 to n, inclusive. The player who says the number p, in this case 1739, wins the game. You can be Player A or B. Which do you choose, and what is your strategy?

Do the initials OJ ring a bell?