SUM OF ALL 3 DIGIT NUMBERS DIVISIBLE BY 7

About the topic "Sum of all 3 digit numbers divisible by 7"

"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

.

First 3 digit number exactly  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"

Last 3 digit number exactly  divisible by 7

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"

Sum of all 3 digit numbers divisible by 7

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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