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 first and also the smallest 3 digit number is 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 last and also the largest 3 digit number is 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 sequence.
And our aim is to find the sum of the terms in the above arithmetic sequence.
Step 2 :
In the arithmetic sequence
105, 112, 119, 126,...................994,
we have
first term = 105
common difference = 7
last term = 994
That is,
a = 105
d = 7
l = 994
Step 3 :
The formula to find the numbers of terms in an arithmetic sequence is given by
n = [(l - a) / d] + 1
Substitute a = 105, l = 994 and d = 7.
n = [(994 - 105) / 7] + 1
n = [889/7] + 1
n = 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 arithmetic sequence is given by
= (n/2)(a + l)
Substitute a = 105, d = 7, l = 994 and n = 128.
= (128/2)(105 + 994)
= 64 x 1099
= 70336
So, the sum of all 3 digit numbers divisible by 7 is 70336.
Note :
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".
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