Re:Math Problems for children from 5 to 15 (2004) [pdf]
by nostoc on Monday July 19, 2021 @03:06PM
(#27886184)
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.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by nbclark on Monday July 19, 2021 @03:33PM
(#27886449)
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)?
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by bentcorner on Monday July 19, 2021 @03:37PM
(#27886488)
I was wondering the same thing but Kopecks are not currently subdivided.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by Arnavion on Monday July 19, 2021 @04:43PM
(#27887055)
The second question's solution requires half-kopeks.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by scotty79 on Monday July 19, 2021 @05:25PM
(#27887430)
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.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by bumbledraven on Monday July 19, 2021 @03:14PM
(#27886250)
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.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by lordnacho on Monday July 19, 2021 @04:00PM
(#27886693)
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.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by bentcorner on Monday July 19, 2021 @03:37PM
(#27886481)
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.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by colinmhayes on Monday July 19, 2021 @03:12PM
(#27886236)
The problem says they combine their money which implies Masha has more than 0 kopeks though.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by zarzavat on Monday July 19, 2021 @10:59PM
(#27889639)
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.
Re:Math Problems for children from 5 to 15 (2004) [pdf]
by parsecs on Monday July 19, 2021 @03:15PM
(#27886259)
> 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..
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.