A remainder problem

If we subtract 35 from the number, then the number has 2009 and 2010 as factors (with no remainder)

The prime factorization of 2009 and 2010,

2009 = 7² * 41

2010 = 2 * 3 * 5 * 67

Observed that they have no common factor so their lowest common multiple is their product:

LCM = 2009 * 2010 = 4038090

So $n$ is equal to some integer multiple of that, plus 35:

$n$ = 4038090k + 35

But 4038090 is evenly divisible by 42, so the remainder will be $35$ .

Given that $n \equiv \begin{cases} 35 \text{ (mod 2009)} \\ 35 \text{ (mod 2010)} \end{cases}$

$\begin{aligned} \implies n & \equiv 2009 \cdot 2010 {\color{#3D99F6}m} + 35 & \small \color{#3D99F6} \text{where }m \text{ is an integer.} \\ & \equiv \boxed{35} \text{ (mod 42)} & \small \color{#3D99F6} \text{Note that } 42 \ | \ 2009 \cdot 2010 \end{aligned}$

Given that $n≡35 \space\ (mod \space\ 2009)$ , we can conclude that $n=2009p+35$ , for $p∈N_0$ . From this we can conclude that $n+7=2009p+42$ . This can be simplified as $n+7=7 \times\ (287p+6)$ , meaning that $7|n+7$ . We can also see that $n≡35 \space\ (mod \space\ 2010)$ , meaning that $n=2010q+35$ for $q∈N_0$ , from which we conclude that $n+7=2010q+42$ , which can further be simplified as $n+7=6 \times\ (335q+6)$ . This means that $6|n+7$ . Because $7|n+7$ , and $6|n+7$ , we can say that $42|n+7$ . From this, we can see that the remainder we get when dividing $n+7$ by $42$ is 35 .

Assume $n=35$ . Therefore $n \bmod{42}\equiv 35\bmod{42}$ .