How to find the number of elements in least common multiple problem? The problem is as follows:

Over a table there is a certain number of muffins, if we count them by
  multiples of four the remainder is three. If we count them by
  multiples of six the remainder is five but if we count them by
  multiples of ten the remainder is nine. What is the least number of
  muffins over the table?

What I tried to do is to build up the equations to get the supposed number as follows:
$\textrm{n is defined as the number of muffins}$
$4k+3=n$
$6k+5=n$
$10k+9=n$
Then i equated those to get the value of $k$.
$4k+3=6k+5$
$2k=-2$
$k=-1$
$10k+9=6k+5$
$4k=-4$
$k=-1$
$4k+3=10k+9$
$6k=-6$
$k=-1$
However in all cases I reach to $-1$ and it does not seem to be reasonable to use it in any of those equations. I'm lost in this one can somebody help me to address this situation the easiest way possible?.
 A: The reason you are getting a nonsensical answer is because there is no reason why the values of $k$ in your three equations should be the same. All you really know is that there are integers $a,b,c$ such that
$$\begin{align}
n &= 3 + 4a, \\
n &= 5 + 6b, \\
n &= 9 + 10c.
\end{align}$$
To solve this, you can iteratively substitute. From the first equation $n = 3 + 4a$, we see that
$$ 3 + 4a = 5 + 6b,$$
or rather that $2a = 1 + 3b = 4 + 3(b-1) = 4 + 3b'$ for some integer $b'$. Note that $b'$ must be even so that the right hand side is even (which must occur as the left hand side is even). So we can write $b' = 2b''$ for some integer $b''$.
Then from $2a = 4 + 6b''$, we have that $a = 2 + 3b''$. Substituting into $n = 3 + 4a$, we find that $n = 3 + 4(2 + 3b'') = 11 + 12b''$ for some integer $b''$. It's annoying to write $b''$, so I'll write $d$ instead of $b''$.
The point is that the information that
$$\begin{align}
n &= 3 + 4a \\
n &= 5 + 6b
\end{align}$$
is contained within the single equation $n = 11 + 12d$. Indeed, if $n = 11 + 12d$ for some integer $d$, then its remainder upon dividing by $4$ is $3$, and its remainder upon dividing by $6$ is $5$. We showed the converse above. Thus from two equations, we have one.
Now we repeat, this time with
$$\begin{align}
n &= 11 + 12 d \\
n &= 9 + 10 c.
\end{align}$$
Equating, we see that $9 + 10c = 11 + 12d$, or rather that $5c = 1 + 6d$. As above, we massage the right hand side a bit and write $5c = 25 + 6(d - 4) = 25 + 6d'$. In this form, we see that $d'$ must be divisible by $5$ to make the right hand side divisible by $5$ (as the left hand side is divisible by $5$). So we can write $d' = 5 d''$ for some integer $d''$. Thus $5c = 25 + 30d''$, or rather $c = 5 + 6d''$. Writing $d''$ is annoying, so I use $e$ instead.
Then we have that $c = 5 + 6e$. Substituting into $n = 9 + 10c$, we have that
$$ n = 9 + 10(5 + 6e) = 59 + 60e.$$
Notice that if $n = 59 + 60e$, then dividing by $4$ leaves remainder $3$, dividing by $6$ leaves remainder $5$, and dividing by $10$ leaves remainder $9$. And to get $n = 59 + 60e$, we just massaged the equations above.
Thus the set of integers that leave the three necessary remainders are exactly those integers of the form $59 + 60e$. The smallest (positive) integer of this form is $59$. (Note that $-1$ is also of this form, and you found $-1$ by essentially guessing a value for $e$ at the start).
So the answer is $59$.

Addendum: the computations I performed were elementary, but they can seem a bit uninspired. But I knew to do this because I was modifying a proof of the Chinese Remainder Theorem, which would apply naively if $\gcd(4,6,10) = 1$ (which it's not). The computations feel more natural when considered in congruences. To see this, let's redo the first pair reduction using congruences.
So $n = 3 + 4a$ is the same as $n \equiv 3 \mod 4$, and $n = 5 + 6b$ is the same as $n \equiv 5 \mod 6$. So we are trying to solve the simultaneous congruence
$$\begin{align}
n &\equiv 3 \mod 4, \\
n &\equiv 5 \mod 6.
\end{align}$$
To do this we write $n = 3 + 4a$ and substitute into the second congruence, getting
$$ 3 + 4a \equiv 5 \pmod 6 \implies 2a \equiv 1 \pmod 3 \implies a \equiv 2 \pmod 3.$$
And thus $a = 2 + 3b$ for some $b$, so that $n = 3 + 4(2 + 3b) = 11 + 12 b$ for some $b$. Or rather $a \equiv 11 \pmod {12}$.
This is exactly as above, except without the various variable shenanigans I employed previously. The rest of the argument works the same.
A: While I cannot compete with mixedmath's more general solution, one interesting aspect of the problem leapt out at me:
$n=4\alpha-1$
$n=6\beta-1$
$n=10\gamma-1$
For $\alpha,\beta,\gamma\in\mathbb{N}$  
So... $n+1=4\alpha=6\beta=10\gamma$ which is to say that $\frac{n+1}{2}$ is divisible by $2, 3$ and $5$.  As the problem asks for the least value of $n$ then we're back on the subject of least common multiple.
$$\text{i.e.}\quad\frac{n+1}{2}=\mathrm{lcm}(2,3,5)=30$$  
However I suspect the problem was structured so as to allow this approach.
