A number theory problem by A Former Brilliant Member

$\begin{aligned} Q & = \frac {1+2+3+\cdots + 1000}{1001} \\ Q & = \frac {1000+999+998+\cdots+1}{1001} \\ Q+Q & = \frac {\overbrace{1001+1001+1001+\cdots + 1001}^{1000 \times 1001}}{1001} \\ 2Q & = 1000 \\ \implies Q & = \frac {1000}2 = \boxed{500} \end{aligned}$

$1 + 2 + 3 + 4 + 5 \ldots \ldots + 1000$ is an Arithmetic Progression, so its sum can be found by using the formula - $\dfrac{n}{2} [2a + d(n-1)]$

Where 'n' is the number of terms, which is $1000$ in this case, 'a' is the first term, $1$ and 'd' is the common difference, which is also $1$ .

Using the formula :

$\begin{aligned} \dfrac{n}{2} [2a + d(n-1)] &= \dfrac{1000}{2} [2(1) + 1(1000-1)] \\ &= 500 \times (2 + 999) \\ &= 500500 \\ &\implies \text{Answer = } \dfrac{500500}{1001} = \boxed{500} \end{aligned}$

$\frac{1 + 2 + 3 + 4 + 5 + \ldots + 1000}{1001} = \frac{(1 + 1000) + (2 + 999) + (3 + 998) + \ldots + (500 + 501)}{1001}$ $= \frac{1001 + 1001 + \ldots + 1001 \mbox{ (500 times)}}{1001} = \frac{1001 \times 500}{1001} = 500$

Formula: $1 + 2 + ... + n = \frac {n(n+1)}{2} \rightarrow 1 + 2 + ... + 1000 = \frac {1000 \times 1001}{2} = \frac {1000}{2} \times 1001 = 500 \times 1001 \rightarrow \frac {1 + 2 + ... + 1000}{1001} = \frac {500 \times 1001}{1001} = \boxed{500}$

1+2+3+...+1000=(1000x1001)/2=500x1001 Don't forget we're dividing by 1001 (500x1001)/1001=500 Therefore the answer is 500