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500 sats \ 3 replies \ @SimpleStacker 9 Dec \ on: Secondary school math puzzle science

Seems like the best way forward is to guess and check. The solution is 6 children and 12 books. That way when they're evenly divided among 6, they each get 2 books (4 less than the number of children.) When divided among 4, the four get 3 books (1 more than they'd otherwise get).

42 sats \ 2 replies \ @Scroogey OP 9 Dec

There is a way without trying, but building the equation was the tricky part (solving it is simple).

Let's define $x$ as the number of children.
Every child should get $x - 4$ books, and there are $x \times (x - 4)$ books.
The number of books freed by the 2 children who don't like them is $2 \times (x - 4)$ .
That's the same number as are given to the remaining children i.e. $(x - 2) \times 1$ .
Hence, $2 \times (x - 4) = (x - 2) \times 1$ or $x = 6$ .

10 sats \ 0 replies \ @unboiled 10 Dec

Interesting. I got there somewhat differently.
x is the number of kids, and y the number of books

y = x * (x - 4), total number of books x kids should get, and
y = (x -2) * (x - 4 + 1), two fewer kids get one more book each for the same total.

ie.

   x * (x-4) = (x-2) * (x-3)
=> x^2 - 4x = x^2 - 5x + 6
=> x = 6

Solve for y to get number of books.

10 sats \ 0 replies \ @SimpleStacker 9 Dec

I initially tried to build two equations with two unknowns, $b$ and $c$ , but they didn't come out linear, so rather than mess with more algebra I resorted to guess and check instead.