How many product?

Since we need to sum the last digits, we should check what they are. For this, we can use modulo arithmetic; specifically, we'll take everything to mod 10.

(I would write $4 \times 5 \equiv_{10} 0$ , but I'll drop all the subscripts). $1 \times 2 \equiv 2, 2 \times 3 \equiv 6, 3 \times 4 \equiv 2, 4 \times 5 \equiv 0, 5 \times 6 \equiv 0,\\6 \times 7 \equiv 2, 7 \times 8 \equiv 6, 8 \times 9 \equiv 2, 9 \times 10 \equiv 0, 10 \times 11 \equiv 0 \\\cdots$ Now, it looks like this pattern repeats with a period of $5$ and an easy check yields this: $(n+5)(n+6) = n^2+11n+30 = n^2 + n + 10(n+3)\\ \equiv n^2 + n = n(n+1)$ We, next, consider the sum of the last digits for each period of $5$ : $2+6+2+0+0 = 10$ Dividing $2010$ by $10$ gives the number of complete periods it would take to sum to $2010$ . $\frac{2010}{10} = 201$ And each period is of length $5$ , so a valid value of $n$ is: $201 \times 5 = 1005$ But... The last two values of $n$ per period contributes nothing to the sum, so we could have stopped at $n=1003$ or $n=1004$ and the sum would have remained unchanged. Clearly, $n=1006$ takes our sum to $2012$ and stopping at $n=1002$ would only give us a sum of $2008$ , so the possible values for $n$ are $\{1003, 1004, 1005\}$ , the sum of which is $\boxed{3012}$ .

Python 3.3:

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n = 1
total = 0
solved = False
while not solved:
    total += int(str(n * (n + 1))[-1])
    if total == 2010:
        answer = 0
        while total == 2010:
            answer += n
            n += 1
            total += int(str(n * (n + 1))[-1])
        solved = True
    n += 1
print("Answer:", answer)