# 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.

- 413
- 692
- 1932
- 2961
- 8882
- 9583
- 16,387
- 23,884
- 49,161
- 71, 645

Answers to previous week’s exercises:

Indicate if the number is divisible or not divisible by 7

- 763 → divisible
- 392 → divisible
- 1444 → not divisible
- 4445 → divisible
- 23331 → divisible
- 38885 → divisible
- 424807 → not divisible
- 133651 → divisible
- 7638476 → not divisible
- 8456728 →divisible