Sum of natural numbers between m and n which is divisible by x

Question

Q.1 The sum of all numbers between 100 and 1000 which is divisible by 13 is .....

Q.2 The sum of all numbers between 200 and 800 which is divisible by 9 is .....

How to solve these type of problems ?

My attempt: Q.2 207+216+...792 =32967 But this is a lengthy process to compute the sum. How to solve these type of problems in shortcut.

Answer 1

$$\sum\limits_{n=0}^{65}(207+9n)=66\times207+9\sum\limits_{n=0}^{65}n=13662+9\dfrac{65\times66}{2}=13662+2145=32967$$

Answer 2

Note that, since we are dealing with an arithmetic progression here,\begin{align}207+216+\cdots+792&=\frac{66\times(207+792)}2\text{ ($66$ is the number of terms)}\\&=33\times999\\&=32\,967.\end{align}

Answer 3

Smallest $3$ digit number divisible by $13$ = $104$
Largest $3$ digit number divisible by $13$ = $988$

Series : $104\,,117\,,130\,, \cdots \,,975\,,988$

This clearly forms an A.P with $a = 104$ , $a_n = 998$ , $d=13$

$$ \begin{align}a_n &= a+ (n-1)d \\ 988 &= 104+(n-1)13 \implies n =69\\ S_n &= \frac n2\left(a+a_N\right) \\ S_n &= \frac{69.(104+988)}{2} \implies S_n = \color{blue}{37674} \end{align}$$

Can you do it for the second part?

Answer 4

Well, yes, there is a shortcut and that is what you are expected to solve.

I'll do question 1:

The first number that that is divisible by $13$ is $13*n>100$ for some $n$.

Figure out what that $n$ is. $100\div 13$ (punch numbers on calculator) $= 7$ with $9$ remainder. So $8*13 =104 > 100$ is the first number that is divisible by $13$.

Now there must be some last number that is divisible by $13$; that is $13* mm < 1000$ for some $m$.

Figure out what that $m$ is. $1000\div 13= 76$ with $12$ remainder. So that last number is $76*13 =988$.

We have to add $8*13 + 9*13+ 10*13 + ........ + 75*13 + 76*13 =$

$13(8+9+10+...... + 75+76)$.

So now we have to add up $8+9 + .....+75 + 76$ and multiply by $13$.

Now.....

Let $K = 8 + 9 + ...........+75 + 76$. Then:

$K = 76 + 75 + ..........+ 9 + 8$ and:

$K + K = (8+76) + (9+75) + ...... + (75+9) + (76+8)$ so:

$2K = 84 + 84 + ...... + 84 + 84 = 84 \times...... $ erk how many terms were there?

Okay. From $1....76$ is $76$ terms. And $1...7$ are $7$ terms... So $8,9....76$ are $76-7=69$ terms.

So

$2K = 69*84$

$K = \frac {69*84}2$

And so $13(7+8 + ..... + 76) = 13K = 13*\frac {69*84}2$.

......

Useful things to learn:

$1 + 2 + 3 + ..... + n = \sum_{k=1}^n k = \frac {n(n+1)}2$.

(If you add up the pairs in ascending and descending order, the pairs add to $n+1$ and there are $n$ pairs so $n(n+1)$ is twice the sum.)

$m + (m+1) + ........ + n = \sum_{k=m}^n k = \frac {(n-m+1)(m+n)}2$.

(If you add up the pairs they add to $m+n$ and there are $n - (m-1)=n-m+1$ pairs.)

And $a*m + a*(m+1) + ..... +a*n = \sum_{k=m}^n ak = a\sum_{k=m}^n k = a\frac {(n-m+1)(m+n)}2$.

.....

So to do question two we are doing

$a*m + a*(m+1) + ........ + a*n$ where $a = 9$ and $9(m-1) < 200 < 9m$ so $m = 23$, and $a*n < 800 < a(n+1)$ so $n = 88$.

So $207 + 216 + ..... + 792 = 9(23+..... + 88) = 9\sum_{k=23}^{88} k = 9\frac {(88-23+1)(88 + 23)}2$