Question

Find the sum of the integers between $100$ and $200$ that are divisible by $9$.

Answer 1

Step 1: Defining the problem

The multiples of $9$ between $100$ and $200$ are $108,117,...,198$

Therefore, this is an arithmetic progression with the first term, $a=108$, common difference, $d=9$ and final term, $a_n=198$

Step 2: Finding number of terms, $n$

An arithmetic progression is defined as, $a_n=a+(n-1)d$

Thus,

$$\begin{gathered} 198=108+(n-1)·9 \\ \Rightarrow n-1=\frac{90}{9} \\ \Rightarrow n=11 \end{gathered}$$

Therefore, number of term, $n=11$

Step 3: Finding the sum of the series

Sum of an arithmetic progression ($S_n$) is given by,

$$\begin{gathered} S_n=\frac{n}{2}\left[2a+(n-1)d\right] \\ =\frac{11}{2}\left[2\times 108+(11-1)·9\right] \\ =\frac{11}{2}\left[216+90\right] \\ =1683 \end{gathered}$$

Therefore, the sum of the multiples of $9$ between $100$ and $200$ is $1683$.

Extra:

Finding first term here,

1. A multiple of $9$ occurs every $9$ number. Thus, there is a multiple of $9$ between $100$ and $109$.
2. Remember that a the sum of the digits of a multiple of $9$ is divisible by $9$.
3. The first two digits in the lower and upper bounds here is $1$ and $0$
4. Thus, the third digit is $9-1-0=8$
5. Thus the first term is $108$

Finding the last term here,

1. Similarly, there is a multiple of $9$ between $200$ and $200-9=191$
2. The first two digits of the upper bound are $1$ and $9$ whose sum is $10$
3. The next multiple of $9$ after $10$ is $18$
4. Thus, the third digit of the upper bound is $18-10=8$