If 
is an ‘n’ digited number, it will be divisible by 7 if 
is divisible by 7.
Proof:
If
So, if
If the number resulted at this step is also high, the procedure can be continued till it can be conveniently decided whether the number is divisible by 7 or not.
Let us take the number 514199 which is actually equal to 7*73457, hence is divisible.
We check the above rule with this number
Step 1: 51419-18=51401
Step 2: 5140-2=5138
Step 3: 513-16=497 which is divisible by 7. Hence the number 514199 is divisible by 7.
Let us try with the number 479833024705. This is NOT divisible by 7
Step 1: 47983302470-10=47983302460
Step 2: 4798330246-0=4798330246
Step 3: 479833024-12=479833012
Step 4: 47983301-4=47983297
Step 5: 4798329-14=4798315
Step 6: 479831-10=479821
Step 7: 47982-2=47980
Step 8: 4798-0=4798
Step 9: 479-16=463 which is not divisible by 7, hence the
number 479833024705 is not divisible by 7
I will update this post with algebraic version to these calculations which makes the process simpler.
I will update this post with algebraic version to these calculations which makes the process simpler.
4 comments:
Thx for the post. I also had one similar method...but the irony is that it would be faster to divide the number by 7and check. But I will certainly share this with my students. Hope there is no exception.
Will you please suggest an easy way to make a child learn times tables of 6..10?
Pratibha
True. In the age of calculators, 'checking' such divisibilities is very easy. But as teachers, we should understand the need of such practices and encourage learners to go through such methods. I think this will help them develop thinking and confidence.
I know one more method.
For any given number to be tested for divisibility of 7, the digits in units, tens, hundreds, thousands, ten thousands and lakhs ……are multiplied by 1, 3, 2, -1, -3, -2 …… respectively in the same order. If the sum of the products is divisible by 7 then the whole number is divisible by 7.
Example: Let us verify, whether 23910551 is divisible by 7 or not?
Units digit, 1 ---> 1 × 1 = 1
Tens digit, 5 ---> 5 × 3 = 15
Hundreds digit, 5 ---> 5 × 2 = 10
Thousands digit, 0 ---> 0 × –1 = 0
Ten Thousands digit, 1 ---> 1 × – 3 = –3
Lakhs digit, 9 ---> 9 × –2 = –18
Ten Lakhs digit, 3 ---> 3 × 1 = 3
Crores digit, 2 --->2 × 3 = 6
--------------------------------------
Adding the products, we get 14, which is divisible by 7. Hence the number, 2,39,10,551 is divisible by 7.
The math behind this method can be understood as follows:
If abc is three digit number, it can be written as 100a+10b+c. If this is to be divisible by 7, then the number c+3b+2a should be divisible by 7. This is true because, (100a+10b+c)-(c+3b+2a)=98a+7b which is divisible by 7.
If the given number is of the form abcdefg (seven digits) we can write it as abcd000+efg or a000000+bcd000+efg and the test can be repeated for each set of three digits except for the left most digits. Some times there will one digit or two digits as the case may be.
Taking -1, -3, -2 is to reduce the size of the resulting number - A good method than what I posted. Thanks to Vidyasagar for sharing.
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