Chinese Remainder Theorem/Simultaneous congruences Find an integer N , 0 ≤ N < 105 such that
N ≡ 2 mod 3,
N ≡ 1 mod 5, and
N ≡ 4 mod 7.
What I have done:
So I have been able to start by splitting up the summation of $x$ into 3 sections:
$ x = 5*7 + 3*7 + 3*5 $
with the first multiplication corresponding with the mod 3 equation, the second multiplication corresponding with the mod 5 equation and the third multiplication corresponding with the mod 7 equation. 
Therefore:
$x = 35 + 21 +15$
However, I know that this isn't complete. But I am not exactly sure on how to proceed.
Any help?

$N \equiv 2 \mod 3$ so $N = 2 + 3a$.
$N \equiv 1 \mod 5$ so $N =1 + 5b$
So $2+3a = 1 + 5b$ so $5b - 3a = 1$.  One solution is $b = 2$ and $a = 3$
So $N \equiv 2 + 3*3 = 1 + 2*5 = 11 \mod 3*5$.
So $N = 11 + 15c$.
$N \equiv 4 \mod 7$ so $N = 4 + 7d$.
So $11 + 15c = 4 + 7d$ so $15c - 7d = -7$.  $c =0$ and $d = 1$ is a solution.
So $N \equiv 11 = 4+7 \equiv 11 \mod 3*5*7 = 105$.
And, indeed, $11 \equiv 2 \mod 3$ and $11 \equiv 1 \mod 5$ and $11\equiv 4 \mod 7$.

Trying to follow your partition reasoning.
$5*7 = 35 \equiv 2 \mod 3$ so that satisfies.
$3*7 = 21 \equiv 1 \mod 5$ so that satisifies.
$3*5 = 15 \equiv 1 \not \equiv 4 \mod 7$ so that does not satisfy.
But $4*3*5 \equiv 4 \mod 7$ so that does.
So $5*7 + 3*7 + 60 = 116$ solves all three.  But $116 > 105$.  But any $k \equiv 116 \mod 105$ so do so $116 - 105 = 11$ will do.
.... or ... when we hae $3*5 \equiv 1 \mod 7$ and we could have figured $4 \equiv -3 \mod 7$ so $-3*3*5 \equiv 4 \mod 7$.
So $N = 5*7 + 3*7 - 3*3*5 = 11$.  (by taking a negative we know we won't get a number too large).
Actually, I had never done the "partitioning" before. 
It works well.  I like it.
