"Sum of all 3 digit numbers divisible by 7" is a difficult problem having had by the students who study math to get prepared for competitive exams.

For some students, getting answer for the questions like "Find the sum of all 3 digit numbers divisible by 7" is never being easy and always it is a challenging one.

Once we know the concept and method of solving, solving the above problem will not be a challenging one.

To get the sum of 3 digit numbers divisible by 7, first we have to find the first and last 3 digit numbers divisible by 7

.

The smallest 3 digit number = 100

The first 3 digit number is also 100.

To find the first 3 digit number divisible by 7, we divide the very first 3 digit number 100 by 7

100/7 = 14.29

We have decimal in the result of 100/7.

Clearly the first 3 digit number 100 is not exactly divisible by 7.

Let us divide the second 3 digit number 101 by 7.

101/7 = 14.43

We have decimal in the result of 101/7 also.

So, the second 3 digit number 101 is also not exactly divisible by 7

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

They are,

**1. Do we have to divide the 3 digit numbers by 7 starting from 100 until we get a 3 digit number which is exactly divisible by 7 ?**

**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 7. **

**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 7, we have to take the very first 3 digit number 100 and divide it by 7.

**Step 2 :**

When we divide 100 by 7 using long division as given above, we get the remainder 2.

**Step 3 :**

Now, the remainder 2 has to be subtracted from the divisor 7.

When we subtract the remainder 2 from the divisor 7, we get the result 5 (That is 7 - 2 = 5).

**Step 4 :**

Now, the result 5 in step 3 to be added to the dividend 100.

When we add 5 to 100, we get 105.

Now, the process is over.

**So, 105 is the first 3 digit number exactly divisible by 7. **

This is how we have to find the first 3 digit number exactly divisible by 7.

**Important Note: **

**This method is not only applicable to find the first 3 digit number exactly divisible by 7. It can be applied to find the first 3 digit number exactly divisible by any number, say "k"**

The largest 3 digit number = 999

The last 3 digit number is also 999

To find the last 3 digit number divisible by 7, we divide the very last 3 digit number 999 by 7.

999/7 = 142.71

We have decimal in the result of 999/7

Clearly the last 3 digit number 999 is not exactly divisible by 7.

Let us divide the preceding 3 digit number 998 by 7.

998/7 = 142.57

We have decimal in the result of 998/7 also.

So, the preceding 3 digit number 998 also is not exactly divisible by 7

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 7 until
we get a 3 digit number which is exactly divisible by 7 ?**

**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 7. **

**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 7, we have to take the very last 3 digit number 999 and divide it by 7.

**Step 2 :**

When we divide 999 by 7 using long division as given above, we get the remainder 5.

**Step 3 :**

Now, the remainder 5 has to be subtracted from the dividend 999.

When we subtract the remainder 5 from the dividend 999, we get the result 994 (That is 999 - 5 = 994).

Now, the process is over.

**So, 994 is the last 3 digit number exactly divisible by 7. **

This is how we have to find the last 3 digit number exactly divisible by 7.

**Important Note: **

**The process of finding the first 3 digit number exactly divisible by 7 and the process of finding the last 3 digit number exactly divisible by 7 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 7. 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 7 in the following steps.

**Step 1 :**

The first 3 digit number divisible by 7 is 105.

After 105, to find the next 3 digit number divisible by 7, we have to add 7 to 105. So the second 3 digit number divisible by 7 is 112.

In this way, to get the succeeding 3 digit numbers divisible by 7, we just have to add 7 as given below.

**105, 112, 119, 126,...............................................994**

Clearly, the above sequence of 3 digit numbers divisible by 7 forms an Arithmetic Progression.

And our aim is to find the sum of the terms in the above A.P

**Step 2 :**

In the A.P 105, 112, 119, .......................................................994 , we have

first term = 105, common difference = 7, last term = 994

That is, **a = 105**, **d = 7** and **l = 994**

**Step 3 :**

The formula to find the numbers of terms in an A.P

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

Plugging a = 105, l = 994 and d = 7

n = [(994-105)/7]+1

n = [889/7]+1 = 127 + 1

n = 128

So, number of 3 digit numbers divisible by 7 is **128**

**Step 4 :**

The formula to find the sum of "n" terms in an A.P is

= **n/2{a+l}**

Plugging a = 105, d = 7, l = 994 and n = 128, we get

= 128/2{105+994}

= 64x1099

= **70336**

**Hence, the sum of all 3 digit numbers divisible by 7 is 70336**

**The method explained above is not only applicable to find the sum of all 3 digit numbers divisible by 7. This same method can be applied to find sum of all 3 digit numbers divisible by any number, say "k".**

When students have the questions like "Find the sum of all 3 digit numbers divisible by 7", in competitive exams, they are stumbling a lot to solve. If we know the way of solving, getting answer for the questions like "Find the sum of all 3 digit numbers divisible by 7" is not a difficult task.

We hope, after having seen the methods and steps explained above, students will not find it difficult to answer the questions like "Find the sum of all 3 digit numbers divisible by 7".

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