bonanova

A better assumption is that the speed you also lose. That leads to a different answer. The train doesn't go at a constant speed.

i think he meant a U-shaped street, wherein both drivers turn in at separate entrances A U-shaped street doesn't provide an intersection. A road radially entering a circular drive provides an intersection - as a longitudinal road intersecting a constant-latitude road very near the poles - or a roundabout - would.

But those aren't the only times.

Oh, c'mon now. The spirit of riddle solving definitely allows for creativity, but assigning a new unit of measure? There's a fine line between being creative and providing a real answer. But I have a feeling he may have been joking. Agree totally. However, if the OP had said "exactly" without the "within 1%", then you'd almost have to stray outside the box. As it stands, I got to learn something new about US currency.

HPT's answer is still the most accurate. Forget the 1% error stuff ... The table is exactly 1 table length long.

What outcome of RPS gives you a winner, a 2nd place winner and a loser? RRR, PPP, SSS and RPS are useless. RRP, PPS and SSR gives two losers RPP, PSS and SRR gives two winners.

If they're chocolate atoms, [shhh!] I will remove the problem by volunteering to eat them.

If opposing cars turn both turn left, their paths may intersect, and the cars may "meet". But not necessarily. Imagine a flagpole at the exact center of the intersection. If they both keep the flagpole on their left as they turn, their paths intersect, but the cars will not collide [meet]. If they both keep the flagpole on their right, their paths do not even intersect.

You're on the right track. If you select just one toy, you know Grinch will keep some of your money. He's priced them to ensure that. If you select two toys they might turn out to be say a $2 toy and a $4 toy, and you could buy them exactly with a $6 coupon. But since the prices have been removed, you don't know this until you get to checkout and then it's too late. If you're going to select just two toys, you'd have to know that every combination of two prices can be paid exactly. So that's what has to be found: if you bring N toys to checkout, there definitely won't be any change for Grinch to keep. What is the smallest value of N?

After disappointing Black Friday sales, management at Grinch's Toy Store found a way to boot profits. Following an approach used in some amusement parks, GTS now sells coupons, which customers then use for payment at checkout. The store's inventory comprises 22 toy models, each clearly marked with a unique, whole-dollar price. Coupons are available in $6, $9 and $20 denominations. Grinch's strategy is threefold. [1] At checkout, extra coupon value is forfeit; no change is given. Example: your total at checkout is $5. You pay with a $6 coupon. Grinch's keeps the $1 change. [2] Prices have been set to ensure there is extra coupon value when toys are bought singly. [3] Each toy model is limited one to a customer. Your task is to visit Grinch's, browse the shelves, and find the minimum number of toys which, when bought together, using coupons, will ensure a fair purchase at checkout. That is, there will be no change for Grinch's to keep. This just in - Grinch's heard you were coming. Fearing your intellectual prowess, they have replaced all price tags with UPC stickers, which only the checkout register can read.

So ... pavvi, what's the answer?

I have a forearm that is about 1 cubit in length. Would that do?

Method #1. Weigh two coins at random. If they don't balance, you've done it in 1 weighing. Yay! Method #2. Determine the smallest number of weighings that guarantees finding the heavy coin. N weighings will distinguish among 3**N possibilities; there are 99 possibilities. 3**4 = 81; 3**5 = 243. Four weighings are not enough; you need 5 to guarantee. Weigh coins of equal number [roughly] one/third, and eliminate [roughly] 2/3 of the coins. You'll guarantee a resolution after the 5th weighing.

20, 82, 100, 4. There are exactly 81 possibilities. Each weighing has 3 distinguishable outcomes. Four weighings can distinguish among 3x3x3x3 = 81 possibilities. Devise a 1st weighing that reduces the number of possibilities to 27 -> 27 against 27; with the 27 suspect balls, Devise a 2nd weighing that reduces the number of possibilities to 9 -> 9 against 9; with the 9 suspect balls, Devise a 3rd weighing that reduces the number of possibilities to 3 -> 3 against 3; with the 3 suspect balls The 4th weighing is 1 against 1; it determines the heavy ball. Four is the fewest weighings, using this method. 4. Three weighings can distinguish among no more than 3x3x3=27 possbilities. With 19 balls that could be heavy or light you have 19x2=38 possibilities. Three weighings won't do it. Use the above method.

Red [scarlet Letter, warning sign, red wine, coals] and blood [running from cleaved flesh, blush, anger's rush] come to mind. The siblings and cousins things are puzzling. Tallest by just a bit could refer to red light having the longest wavelength of visible light. Except the "by just a bit" couldn't mean percentage - red/blue =~ 1.75 wavelength ratio - almost double. Perhaps erythrocytes [red blood cells] with and without oxygen - two siblings. Or the siblings could be the other two types of blood cells -- thrombocytes [platelets] and leukocytes [white blood cells]. Cousins could be other red cells. But that's more poetic than factual.

Another approach is to make two piles with 0 pennies in each pile. Each pile will have 0 pennies with heads showing. *scurrying back inside the box*

If A=1 then ABCDEF = BCDEF

$16,688 if you include collectors denominations. The $100,000 Gold Certificate was never released into general circulation and was only used in fiscal channels. This note cannot be legally held by currency note collectors. If it were, the amount would have been $116,688 The highest circulated denomination is now $100. So you probably got only $188.

Hey, not to worry ... You might want to add some conditions, or start fresh with a new puzzle. As for wasting anyone's time ... that just might be why many of us drop by ... No one's making a salary here.

Well, all rational numbers do have exact representations as integer ratios; but not as finite-length decimals. But irrational numbers [pi, e.g.] lack exact representations both ways. pi ~= 22/7 pi ~= 3.1415926536 ... Now, although the precision of both representations can be improved to an arbitrary level, it's much easier to add a digit to the decimal representation than it is to find the next-more-precise integer ratio. Also, just as 1/3 may look more compact than 0.333333 ... one can say .3 looks more compact than 1/0.33333 ... But to be fair .3 and 3/10 are about the same. Your comment is interesting but not generally true.