The divisibility test for 7 as taught in schools

Last week we discussed, using 7 as an example, a divisibility test which can be used for any divisor. We will now discuss the divisibility rule for 7 as commonly taught in schools: “the difference between twice the unit’s digit of a number and the remaining part of that number, must be divisible by 7 or zero”.

For example, 616 is divisible by 7 since 61 – (2 x 6) = 61 – 12 = 49 which is a multiple of 7.

For 6461, we have 646 – (2 x 1) = 644. We repeat the operation again: 64 – (2 x 4) = 56. We can say that 6461 is divisible by 7.  

If we try this rule on 16,414, weget 1641 – (2 x 4) = 1633; 163 – (2 x 3)= 157; 15 – (2 x 7) = 1. Then 16,414 is not divisible by 7.

This rule is a specific application of the general “By Addition and By Subtraction” sutra we discussed last week. And we believe that instead of teaching this specific rule for 7 in schools, the general method should be explained instead.

By Subtraction of known multiples of a divisor, we can create zeroes at the end of a number. For 7, we can conveniently use 21, for reasons which will become obvious as we go on. Using our earlier examples, we can subtract 21 from 6461 to get 6440. We can then delete the ending zero, get 644 and continue. To convert the ending digit 4 into a 0, we must subtract 4 x 21 or 84 from 644. 644 – 84 = 560. If we disregard the ending 0, we will get 56 which is clearly a multiple of 7.

In Vedic Math, this systematic application of “By Division” is called negative osculation with 2 as the negative osculator of 7.

Now if we try it on 3612, we get 361 – (2 x 2) = 357 and 35 – (2 x 7) = 21. Here we can say that 3612 is divisible not only by 7 but also by 21.

Now we can try it on 17. Since 3 x 17 = 51, we can use 5 as a negative osculator for 17.  To test if 1411 is divisible by 17, we have 141 – (5 x 1) = 136, 13 – (5 x 6) = – 17. Divisible by 17.

For 2091, we have 209 – (5 x 1) = 204; 20 – (5 x 4) = 0. Divisible by 17 and 51.

Keeping in mind that composite numbers must tested for divisibility for its factors and since the divisibility rules for common numbers such as 2, 3, 4, 5, 8 and 9, we often need only to develop rules for prime numbers ending in 1, 3, 7 and 9.

We have seen how negative osculation  was used for 7 and 21; and 17 and 51, we can convert all those prime numbers into numbers ending in 1 by appropriate multiplication, i. e. 13 x 7 = 91 and use 9 as the negative osculator of 13.

Practice exercises: Identify which of the following numbers are divisible by 7, and by 21.

1. 413
2. 692
3. 1932
4. 2961
5. 8882
6. 9583
7. 16,387
8. 23,884
9. 49,161
10. 71, 645

Answers to previous week’s exercises:

Indicate if the number is divisible or not divisible by 7

1. 763 → divisible
2. 392 → divisible
3. 1444 → not divisible
4. 4445 → divisible
5. 23331 → divisible
6. 38885 → divisible
7. 424807 → not divisible
8. 133651 → divisible
9. 7638476 → not divisible
10.  8456728 →divisible