1. The problem statement, all variables and given/known data
A factory has gotten an order for $N$ products. For this job it has $n$ available machines.If all the machines would work simultaneously the order would have been completed in 24 hours. However due some complications there exist only two possible way to build the products:
$a)$ The machines begin working one after the other like this:
In the first hour only one machines works In the second hour two machines work 3 hour tree machines are working
$\vdots$
n-th hour and all next hours n machines are working
$b)$ $n-5$ machines are working all the time
in case $b)$ the products are made 6 hours faster than in case $a)$ How many machines are in the factory
2. The attempt at a solution
I was asked this question by some students I was helping out and I was really surprised when I was unable to answer it . The way I tried to tackle this was the following:
First I tried to calculate the amount of work 1 machine does in 1 hour . I know that $n$ machines complete $N$ in 24 hours therefore one machines does $\frac{N}{n}$ of work in 24 hours. therefore the amount of work a machine does in 1 hour should be $\frac{N}{24n}$( not sure if this is correct)
then I started with case $4a)$
1 hour->1 machine->$\frac{N}{24n}$ of total work done
2 hour ->2 machines->$\frac{N}{24n}+\frac{2N}{24n}$ of total work done
3 hour -> 3 machines ->$\frac{N}{24n}+\frac{2N}{24n}+\frac{3N}{24n}$ of total work done
here is where I started noticing a recursion
n-th hour ->n machines ->$\sum\limits_{i=1}^{n} \frac{i \cdot N}{24n}=\frac{N}{24n}(1+2+3+4+\cdots+n)=\frac{N}{24n}\frac{n(n+1)}{2}=\frac{N(n+1)}{48}$
and every next hour we would add another $\frac{N}{24}$ of work
then I tried using case $b)$
we know that every hour we get $\frac{(n-5)\cdots N}{24n}$ work done
1 hour ->$\frac{(n-5)\cdot N}{24n}$ work done
2 hour -> $\frac{(n-5)\cdot N}{24n}+\frac{2(n-5)\cdot N}{24n}$
3 hour-> $\frac{(n-5)\cdot N}{24n}+\frac{2(n-5)\cdot N}{24n}+\frac{3(n-5)\cdot N}{24n}$
$\vdots$
t-th hour $\frac{(n-5)\cdot N}{24n}\sum\limits_{i=1}^{t} i=\frac{(n-5)\cdot N}{24n}\frac{t(t+1)}{2}$
however after (if it even is correct ) I'm not completely sure how to continue: I know I still haven't used that case $b)$ takes 6 less hours to complete than $a)$ but i'm now exactly sure how to use that here. I know there is something obvious that i'm missing but I just cant find it
Any help is greatly appreciated
A simple way to tackle this interesting word puzzle is to work in machine-hours of which $24n$ are needed. Let (a) take $T$ hours. This gives the following equations. $$nT-\frac{1}{2}n(n-1)=24n$$ $$(T-6)(n-5)=24n$$ Eliminating $T$ you will obtain a quadratic in $n$ with solution $$n=25$$(Then $T=36.$)