Sick-quences (1)

Note that digital sum of an integer $N> 0$ is

• $N (mod 9)$ if 9 does not divide $N$
• 9 if 9 divides $N$ .

However, note that the one of the 11 consecutive integers $r, r+1, ..., r+10$ must always be a multiple of 9, hence their product is a multiple of 9. Thus the sum in the question is a multiple of 9 i.e. the answer is 9. (It is not necessary to calculate the actual value of the sum).

Well, overrated! (It is level 5 now)

$\sum _{ r=1 }^{ 11 }{ r(r+1)(r+2)....(r+9)(r+10) }$

${T}_{r} = \frac{(r+10)!}{(r-1)!}$

${T}_{r} = 11!\left( \begin{matrix} r+10 \\ 11 \end{matrix} \right)$

$11!\sum_{ r=1 }^{ 11 }{\left( \begin{matrix} r+10 \\ 11 \end{matrix} \right)}$

Now using the formula $\left( \begin{matrix} r \\ r \end{matrix} \right) \quad +\quad \left( \begin{matrix} r+1 \\ r \end{matrix} \right) \quad +\quad \left( \begin{matrix} r+2 \\ r \end{matrix} \right) \quad +....\quad +\quad \left( \begin{matrix} n \\ r \end{matrix} \right) \quad \quad =\quad \left( \begin{matrix} n+1 \\ r+1 \end{matrix} \right)$ ,

$\sum_{ r=1 }^{ 11 }{\left( \begin{matrix} r+10 \\ 11 \end{matrix} \right)} = \left( \begin{matrix} 22 \\ 12 \end{matrix} \right)$

Now the sum becomes -

$\sum _{ r=1 }^{ 11 }{ r(r+1)(r+2)....(r+9)(r+10) } = 11!\left( \begin{matrix} 22 \\ 12 \end{matrix} \right)$

Now finding the above(which is easy) and then finding digital sum, Io, you get the answer as $\boxed{9}$

Let the summation be written as $S=\sum_{r=1}^11 T_r$

where $T_r =\prod_{j=0}^11 (r+j)$

$T_r$ is a product of 11 consecutive integers. One of those integers would be divisible by 9. Hence, $T_r$ would be divisible by 9.

Since, $S$ is a summation of terms, each of which is a multiple of 9, $S$ is also divisible by 9.

Applying the rules of divisibility by 9, the digit sum should come out to be 9.