To get the sum of 3 digit numbers divisible by 8, first we have to find the first and last 3 digit numbers divisible by 8.

The first and the smallest 3 digit number is 100.

To find the first 3 digit number divisible by 8, we divide the very first 3 digit number 100 by 8.

100/8 = 12.5

We have decimal in the result of 100/8.

Clearly the first 3 digit number 100 is not exactly divisible by 8.

Let us divide the second 3 digit number 101 by 8.

101/8 = 12.625

We have decimal in the result of 101/8 also.

So, the second 3 digit number 101 is also not exactly divisible by 8

Here, students may have some questions on the above process.

They are,

1. Do we have to divide the 3 digit numbers by 8 starting from 100 until we get a 3 digit number which is exactly divisible by 8 ?

2. Will it not take a long process ?

3. Is there any shortcut instead of dividing the 3 digit numbers 100, 101, 102.... one by one ?

There is only one answer for all the above three questions.

That is, there is a shortcut to find the first three digit number which is exactly divisible by 8.

**SHORTCUT**

What has been done in the above shortcut ?

The process which has been done in the above shortcut has been explained clearly in the following steps.

**Step 1 :**

To get the first 3 digit number divisible by 8, we have to take the very first 3 digit number 100 and divide it by 8.

**Step 2 :**

When we divide 100 by 8 using long division as given above, we get the remainder 4.

**Step 3 :**

Now, the remainder 4 has to be subtracted from the divisor 8.

When we subtract the remainder 4 from the divisor 8, we get the result 4 (That is 8 - 4 = 4).

**Step 4 :**

Now, the result 4 in step 3 to be added to the dividend 100.

When we add 4 to 100, we get 104.

Now, the process is over.

So, 104 is the first 3 digit number exactly divisible by 8.

This is how we have to find the first 3 digit number exactly divisible by 8.

**Important Note : **

This method is not only applicable to find the first 3 digit number exactly divisible by 8. It can be applied to find the first 3 digit number exactly divisible by any number, say "k"

The last and the largest 3 digit number is 999.

To find the last 3 digit number divisible by 8, we divide the very last 3 digit number 999 by 8.

999/8 = 124.875

We have decimal in the result of 999/8.

Clearly the last 3 digit number 999 is not exactly divisible by 8.

Let us divide the preceding 3 digit number 998 by 8.

998/8 = 124.75

We have decimal in the result of 998/8 also.

So, the preceding 3 digit number 998 also is not exactly divisible by 8

Here, students may have some questions on the above process.

They are,

1. Do we have to divide the 3 digit numbers .......997, 998, 999 by 8 until we get a 3 digit number which is exactly divisible by 8 ?

2. Will it not take a long process ?

3. Is there any shortcut instead of dividing the 3 digit numbers ...........997, 998, 999 one by one ?

There is only one answer for all the above three questions.

That is, there is a shortcut to find the last three digit number which is exactly divisible by 8.

**SHORTCUT**

What has been done in the above shortcut ?

The process which has been done in the above shortcut has been explained clearly in the following steps.

**Step 1 :**

To get the last 3 digit number divisible by 8, we have to take the very last 3 digit number 999 and divide it by 8.

**Step 2 :**

When we divide 999 by 8 using long division as given above, we get the remainder 7.

**Step 3 :**

Now, the remainder 7 has to be subtracted from the dividend 999.

When we subtract the remainder 7 from the dividend 999, we get the result 992 (That is 999 - 7 = 992).

Now, the process is over.

So, 992 is the last 3 digit number exactly divisible by 8.

This is how we have to find the last 3 digit number exactly divisible by 8.

Important Note :

The process of finding the first 3 digit number exactly divisible by 8 and the process of finding the last 3 digit number exactly divisible by 8 are completely different.

Be careful! Both are not same.

The methods explained above are not only applicable to find the first 3 digit number and last 3 digit number exactly divisible by 8. They can be applied to find the first 3 digit number and last 3 digit number exactly divisible by any number, say "k"

Let us see how to find the sum of all 3 digit numbers divisible by 8 in the following steps.

**Step 1 :**

The first 3 digit number divisible by 8 is 104.

After 104, to find the next 3 digit number divisible by 8, we have to add 8 to 104. So the second 3 digit number divisible by 8 is 112.

In this way, to get the succeeding 3 digit numbers divisible by 8, we just have to add 8 as given below.

104, 112, 120, 128,...............................992

Clearly, the above sequence of 3 digit numbers divisible by 8 forms an arithmetic sequence.

And our aim is to find the sum of the terms in the above arithmetic sequence.

**Step 2 :**

In the arithmetic sequence

104, 112, 120, 128,...............................992,

we have

first term = 104

common difference = 8

last term = 992

That is,

a = 104

d = 8

l = 992

**Step 3 :**

The formula to find the numbers of terms in an arithmetic sequence is given by

n = [(l - a) / d] + 1

Substitute a = 104, l = 992 and d = 8.

n = [(992 - 104) / 8] + 1

n = [888/8] + 1

n = 111 + 1

n = 112

So, number of 3 digit numbers divisible by 8 is 112.

**Step 4 :**

The formula to find the sum of 'n' terms in an arithmetic sequence is given by

= (n/2)(a + l)

Substitute a = 104, d = 8, l = 992 and n = 112.

= (112/2)(104 + 992)

= 56 x 1096

= 61376

So, the sum of all 3 digit numbers divisible by 8 is 61376.

**Note :**

The method explained above is not only applicable to find the sum of all 3 digit numbers divisible by 8. This same method can be applied to find sum of all 3 digit numbers divisible by any number, say "k".

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