Let

$$
R=3 \times 9+4 \times 10+5 \times 11+\cdots+2003 \times 2009
$$

$$
S=1 \times 11+2 \times 12+3 \times 13+\cdots+2001 \times 2011
$$

(a) Which is the larger number: $\mathrm{R}$ or $S$?

(b) Calculate the difference between the larger and the smaller.
(a) Each term of $S$ is of the form $n \times(n+10)=n^{2}+10 n$ and each term of $R$ is of the form $(n+2) \times(n+8)=n^{2}+10 n+16$ with $\mathrm{n}=\{1,2, \ldots, 2001\}$ in both cases. Thus, for every n, each term of $R$ is greater than the corresponding one in $S$, making $\mathrm{R}>\mathrm{S}$.

(b) The difference between the corresponding terms is equal to

$$
\left(n^{2}+10 n+16\right)-\left(n^{2}+10 n\right)=16
$$

Since there are 2001 terms, the difference between R and S is equal to $16 \times 2001=32016$.
Helpful Facts: The sum of the digits of a multiple of 9 is divisible by 9.
Hint: Observe that each term of $S$ is of the form

$$
n \times(n+10)
$$

and each term of R is of the form

$$
(n+2) \times(n+8)
$$

## Helpful Facts:

$$
\begin{aligned}
& (a+b) \times(c+d)= \\
& a c+a d+b c+b d
\end{aligned}
$$

Hint: In item (b), analyze the numbers that have the sum of their digits greater than or equal to 17.
Hint: For four consecutive numbers, use the notation $x, x+1$, $x+2, x+3$

Helpful Facts: (a) The only even prime number is 2. (b) The only prime number that is a multiple of 3 is 3.