> A brick weighs one pound and half the brick. How many pounds does the brick weigh?

I am a native english speaker and am having a hard time parsing this one. The only sane interpretation I can think of is that one pound + half the brick = the whole brick.

EDIT: I think the reason it is so confusing to me is because "and half the brick" sounds like (the start of) an independent thought. "A brick weighs one pound and half the brick was painted yellow".

This version is much clearer, IMO: "A brick weighs one pound plus half a brick". Maybe there is a fear that wording the problems too clearly makes the solution obvious.

Correct.

These problems are worded to be deliberately confusing, especially #1. Is it a translation issue or are they worded because the math itself is too obvious once the wording has been deciphered?

Here’s a way to use algebra to grind out the solution to #1 with no particular insight needed.

Assume that prices and the amount of kopeks a person has are both represented by nonnegative integers. Let A be Masha’s kopeks, let I be Misha’s kopeks, and let B be the price of the book. We are given the following:

A = B - 7 (1)

I = B - 1 (2)

A + I < B (3)

Substituting (1) and (2) into (3) yields

B - 7 + B - 1 < B (4)

This simplifies to

B < 8 (5)

B = 7 satisfies (5) and, from (1) and (2), implies that A = 0 and I = 6, which together satisfy the givens (1), (2), and (3). So B = 7 is a solution. Furthermore, we cannot have B < 7 or else (1) would imply A < 0, contradicting the assumption that our variables are represented by nonnegative integers. So B = 7 is the only solution.

It's also quite easy to do intuitively.

Since Misha is missing only one kopeck, had Masha owned any amount, the sum would have been enough to buy the book.

Therefore, Misha doesn't have any money, and the price book is what Misha is missing : 7 kopecks.

Couldn't the book cost 7.5k and one has 6.5 and the other has 0.5? Along those lines, isn't anything in the range of costing 7->8 (non-inclusive) acceptable (e.g. 0.9k and 6.9k)?

I was wondering the same thing but Kopecks are not currently subdivided.

https://en.wikipedia.org/wiki/Kopek

The second question's solution requires half-kopeks.

These problems aren't current or modernized. There's no way a bottle with a cork in problem #2 costs 10 kopek.

current 10 kopek is worth 0.0013$

They probably come from time when 0.5 kopek was the smallest coin.

#1 problem doesn't have single non-zero solution if the smallest coin has larger or smaller.

That’s a nice approach. (It’s the same one given by bencollier earlier: https://news.ycombinator.com/item?id=27885681). I regard it as requiring a bit of insight, as opposed to my approach, which is more like grinding gears to reach a conclusion.

This is one of my main observations growing up with math: it's the moments of beauty and elegance that are the most exciting, but the grinding gears thing is also a necessity. They complement each other. For instance when you're just learning the basics there's a lot of these "wow what an insight" but over time you figure out that people have distilled it into a mechanical procedure, which also has some attraction to it. Something like quadratic equation turns the search for a pair of numbers that add up some one thing and multiply to another into a simple formula. You then use that mechanism to build ever more elaborate ones.

I came to the same conclusion the same way but it felt wrong due to the phrase "They combined their money to buy one book to share". Perhaps the phrase lost something in translation.

The problem says they combine their money which implies Masha has more than 0 kopeks though.

Yes this is the flaw in the question. They should have said that the sum of their money is insufficient. Combine implies a physical action which can't happen if one of the parties has nothing.

> They combined their money to buy one book to share, but even then they did not have enough.

Is there some error here? I read it as "even then they did not have enough individually"

I used b-7+b-1=b to arrive at b=8... My math skills are pretty awful so I can't say if this even makes sense..

What is your objection to question #13?

I suppose the question doesn’t mention that volume 1 is on the left and volume 2 is on the right but I guess that would be assumed by any speakers of left to right languages.

> I suppose the question doesn’t mention that volume 1 is on the left and volume 2 is on the right but I guess that would be assumed by any speakers of left to right languages.

The answer is supposed to be 4mm; the only way for that to work if volume 1 is on the left is for the bookworm to gnaw its way out of the book from v.1.p.1, cross the outside of the two books without gnawing anything until it reaches the back cover of volume 2, and then gnaw its way through that cover to reach the final page of volume 2.

I don't think that's what the question has in mind. The point of being a bookworm is that you don't leave the book. So the answer would appear to require that volume 2 is shelved in front of volume 1. I don't know why that would be the case.

Diagram for the desired solution, on the shelf:

  V1 V2

In that order you can see that the pages in each book are in this order:

  |    V1    |    V2    |
  +----------+----------+
  |9876543210|9876543210|

(0-indexing of the pages for fun, plus it fit better, also reminded me of annoying protocol specs that mix 0- and 1-based indices with different elements)

The first page of V1 is the rightmost page (shelved) of Volume 1, and the last page of V2 is the leftmost page (shelved) of Volume 2. So the bookworm ends up going only through the covers. Having volumes shelved in order from left-to-right is conventional in left-to-right languages since that's the same direction we read, and you'd want to "read" through the titles to find the volume you wanted.

This is incorrect. You seem to be making the same error as the author says the editors made in the footnote at the bottom.

If volume 1 is on the left, and the worm goes from page 1 of volume 1 to the last page of volume 2, it travels 4mm in a straight line.

Page 1 of volume 1 and the last page of volume 2 will be right next to each other, if volume 1 is on the left.

Wow, this relies on both books being in the same orientation, with front cover to the right. It assumes a lot. For perspective, I for years kept books shelved upside down because that orientation was easier for me when reading spines.

I guess it relies on the books being ordered and arranged the same way they'd be in every single bookstore and library in the world (in left-to-right ordering countries).

But it's true, maybe this is too much to assume. Most of the time when I've seen this puzzle it's shown the book spines in an image to make it clear, and many people still can't get it. Then again, that would rely on knowing whether it was using top-to-bottom or bottom-to-top book title orientation, so perhaps the only solution is for the author to spell out "the first page of volume 1 is next to the last page of volume 2."

> in every single bookstore and library in the world

Not so fast. Some cultures (Japan for one, I think China and Taiwan as well?) have page-ordering right-to-left but books are generally stacked left-to-right from what I've seen (and in ether case bookstores don't order volumes differently depending on if it's native right-to-left books or foreign right-to-left ones).

German books have the orientation of the writing on the spine flipped. I don't like storing books upside down, so it makes a mess in my mixed English and German bookshelf.

There's a language-wide order in Germany of direction the of writing on the spines of a books?

I just checked my bookshelf and my books go both ways.

Ah. I just checked, and while all the English ones seem to have a consistent orientation, the German ones indeed don't. Never noticed, huh...

The point with this question is that if volume 1 is on the left and volume 2 is on the right, the first page of volume 1 is facing right and the last page of volume 2 is facing left, so the only thing between them is the two covers. Hence, the answer is 4 mm.

It also assumes that the pages aren't facing out.

could also be the case that the two volumes are empty, ie have no pages

The problem statement gives us that there are 2cm of pages in each book. So they are not empty. The confusion is in which order the books would be on the shelf, and consequently which direction the bookworm would be moving and through what.

what book has no pages? Also the problems states:

> The pages of each volume are 2 cm thick

The pages would be less than 20mm thick in that case.

> side-by-side on a bookshelf

Seems pretty explicit to me

Assuming volume 1 is on the left and the books have their spine facing out.

Reasonable assumptions, but relying on implicit knowledge nonetheless.

Which volume is to the left? Thus the ambiguity.

> A bookworm has gnawed through (perpendicular to the pages) from the first page of volume 1 to the last page of volume 2.

How would a worm eat perpendicular to the pages and go from the first page of volume 1 to the last page of volume 2?

It can only be vol1->vol2

> How would a worm eat perpendicular to the pages and go from the first page of volume 1 to the last page of volume 2?

Sorry, what is the question supposed to be? You're positing a contradiction between two facts:

- The worm's path is perpendicular to the pages.

- The worm's path begins at the first page of volume 1, and ends at the final page of volume 2.

What's the contradiction?

The bookworm could have moved right to left from volume 1 to volume 2. Assuming the spines are facing out, the books are right-way up, and volume 2 is on the left of volume 1, then the answer would be 44mm.

While the answer does assume that V1 is on the left, there's no contradiction in your statement. If V2 happened to be on the left, it would still be perfectly logical for "a worm eat perpendicular to the pages and go from the first page of volume 1 to the last page of volume 2." They would simply have to go through more pages.

My answer to #1 is less than 8 kopecks, and Masha has less than one. There's a problem: nowadays kopecks are the minimal unit of currency. Either it means you have to think of the old Imperial money units (polushka, 1/4 of kopeck), or think of fractional amounts of money.

I assumed kopecks were pennies. If Misha needs one, and Masha doesn't have enough to give her one, than Masha must have none. So the answer follows from that.

How can they combine their money if one of them doesn't have any?

Communism?

Ok. I literally laughed out loud.

But naturally, at first you assume the numbers are natural. :)

Well, yes, otherwise you'd have an infinite number of solutions.

Seems like Mishalacked isn't great at making deals.

1. Depends on whether kopecks are divisible into a smaller monetary unit or not. If they are divisible into 100 units, I believe the answer is "anywhere between 7.00 and 7.99 kopecks".

(Problem #2 requires kopecks to be divisible.)

I think part of the point of this brochure is to think about the problems intuitively in the context they are presented. So in the first problem it's just kids trying to buy their first book, it would be silly to think Masha had a fraction of a kopeck (assuming you understand what a kopeck is, I really think it should have been translated as cent) and that the answer could be in range [7, 8). This may be what he talks about when he says that many academics fail at these problems.

Similarly, in problem #2 the cork indeed costs 0.5 kopecks but in this case we're just thinking about cost conceptually, not in terms of how much money a person actually has on hand.

Indeed, but it likewise seems intuitively reasonable to think that a book costs much more than 7 cents (or 7 times whatever the atomic unit of currency is) and that over 100 times the atomic unit is more reasonable.

Before 1917, they were divisible into 4 polushkas.

They're not. It's like a penny. Sure you may have parts of cents like with gas, but for girls buying books, it's the lowest currency value.

Long ago, many currencies had more divisions than they do now.

The penny of GBP came in quarter pennies (farthings) until 1950.

Half US-cent coins were made until 1857.

> (Problem #2 requires kopecks to be divisible.)

Does it? Did I fail at problem 2? I got:

Bottle + cork = 10

Bottle = 9cork

bottle/9 = cork

9(bottle + bottle/9) = 9(10)

9 bottle + bottle = 90

10 bottle = 90

bottle = 9

9 = 9cork

1 = cork

I read the second statement ("the bottle itself is 9 kopecks more expensive than the cork.") as:

Bottle = 9 + Cork

Your statement (Bottle = 9 * Cork) would be "the bottle is 9 times as expensive as the cork".

I solve it to:

Bottle = 9.5 Kopecks.

Cork = 0.5 Kopecks.

Works even if Kopecks are divisible: say the books value is 7.5, Mash must have 0.5 (7.5 - 7) and Misha 6.5 (7.5 - 1), however now when you combine them they sum to exactly 7.5 not less than, the only way to arrive at less than is if Masha has 0. So its always exactly 7

6.5 + 0.5 is 7.0, not 7.5, so that should be valid.

The book's cost can lie anywhere between [7, 8).

Right on the original typo, but still not convinced, I’ve rephrased to original to try to be clearer

> say the books value is 7.5, Mash must have 1.5 (7.5 - 7) and Misha 6.5 (7.5 - 1), however now when you combine them they sum to exactly 7.5 not less than

There are several problems with this:

- 7.5 - 7 is 0.5, not 1.5

- 1.5 + 6.5 is 8, not 7.5

- 0.5 + 6.5 is still less than 7.5

The problem specifies that 2x - 8 < x. There is no way to constrain this to the specific solution x = 7. Everything would work fine if the book cost -2.6 kopecks.

Thank you :) I’m being very clueless today

6.5 plus 0.5 is 7.0, they don't add up to exactly 7.5.

A kopeck is like a cent (1/100th of a ruble).