Divisibility Rules for 7

Divisibility rules are one of the important topics of study in school mathematics, especially in upper primary classes They enable us to quickly identify if one number is divisible by another. We know various methods for checking the divisibility of a number by 2, 3, 4, 5, 6, 8, 9, 10 etc. It is also clear that checking of divisibility of the given number by some numbers is quite easy, while for some numbers is a bit complicated.

Divisibility by 7 is a challenging one, with many attempts made to simplify the rule. Chika’s divisibility rule for 7 is a recent one among them. Here, we shall discuss three different divisibility methods for 7, using existing methods which add new dimensions to the concept.

Method 1: Doubling the unit digit

With the above examples, we understand that this method is useful for checking divisibility by 7 without performing long division for a 3-digit number, but is quite lengthy for 4 or more-digit numbers.

Justification of the rule

Suppose \(N = 1000\ a_{3}+ 100\ a_{2} + 10\ a_{1} + a_{0}\)
(Where \(a_{0}, a_{1}, a_{2}, a_{3}\) are the digits of the 4-digit number N)

According to the rule, we write the truncated version (say \(N_{T}\)) without the unit digit of N and then take away (subtract) from \(N_{T}\) twice the unit digit to get a new number (say M).

\(N_{T} = 100\ a_{3}+ 10\ a_{2} + a_{1}\) (Note the change in the place values after the number is truncated)
\(M = N_{T} – 2a_{0} = 100\ a_{3} + 10\ a_{2} + a_{1} – 2a_{0}\)

Our rule says that if M is a multiple of 7, then N is also a multiple of 7.
Assume that M is a multiple of 7, i.e. M = 7k for some whole number k.

Then, \(M = 7k = 100 a_{3} + 10 a_{2} + a_{1} – 2a_{0}\) or \(100 a_{3} + 10 a_{2} + a_{1} =7k + 2a_{0}.\) Substituting this in N, we get

\(N = 1000\ a_{3} + 100\ a_{2} + 10\ a_{1} + a_{0}\)
\(N = (1000\ a_{3} + 100\ a_{2} + 10\ a_{1} )+ a_{0} = 10(100\ a_{3} + 10\ a_{2} + a_{1} ) + a_{0}\)
\(= 10(7k + 2a_{0}) + a_{0} = 70k + 21a_{0} = 7(10k + 3a_{0})\)

So, if M is a multiple of 7, then so is N.
This can easily be generalized to any number of digits.

Method 2: Multiplying the unit digit by 5

One may provide a justification, which is very similar to the previous one as follows.

Suppose \(N = 1000\ a_{3} + 100\ a_{2}+ 10\ a_{1} + a_{0}\)
(Where \(a_{0}, a_{1}, a_{2}, a_{3}\) are the digits of the 4-digit number N)

According to the rule, we write the truncated version (say NT) of N and add five times the unit digit to get a new number (say M).

\(N_{T} = 100\ a_{3}+ 10\ a_{2} + a_{1}\)
\(M = N_{T} + 5a_{0} = 100\ a_{3} + 10\ a_{2} + a_{1} + 5a_{0}\)

Our rule says that if M is a multiple of 7, then N is also a multiple of 7.
Assume that M is a multiple of 7, i.e. M = 7k for some whole number k.

Then, \(M = 7k = 100 a_{3} + 10 a_{2} + a_{1} + 5a_{0}\) or \(100 a_{3} + 10 a_{2} + a_{1} =7k – 5a_{0}\). Substituting this in N, we get

\(N = 1000\ a_{3} + 100\ a_{2} + 10\ a_{1} + a_{0}\)
\(N = (1000\ a_{3} + 100\ a_{2} + 10\ a_{1} )+ a_{0} = 10(100\ a_{3} + 10\ a_{2} + a_{1} ) + a_{0}\)
\(= 10(7k – 5a_{0}) + a_{0} = 70k – 49a_{0} = 7(10k – 7a_{0})\)

So, if M is a multiple of 7, then so is N.
This can easily be generalized to any number of digits.

Method 3: Grouping of digits (Rule – 1-3-2)

This is yet another way of checking for divisibility by 7. Let’s illustrate it step by step.

1. Starting from the unit place of the number, make groups of three digits. The last group will contain the remaining digits.
2. In each group, multiply the right-most digit by 1, the next by 3 and the left-most by 2.
3. Add all the products obtained in each group.
4. Find the sums of the odd and even-numbered groups.
5. If the difference of these two sums is divisible by 7 or is 0, then the original number will be divisible by 7.

Justification of the rule

Suppose \(N = 100000\ a_{5} + 10000\ a_{4} + 1000\ a_{3} + 100\ a_{2} + 10\ a_{1} + a_{0}\)
(Where \(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5},\) are the digits of the 6-digit number N)

\(S1 = a_{0} \times 1 + a_{1} \times 3 + a_{2} \times 2\)
\(S2 = a_{3} \times 1 + a_{4} \times 3 + a_{5} \times 2\)
\(M = S_{1} – S_{2}\)

Our rule says that if M is a multiple of 7, then N is also a multiple of 7.
Assume that M is a multiple of 7, i.e. M = 7k for some whole number k.

\(7k = (a_{0} × 1 + a_{1}, × 3 + a_{2} × 2) – ( a_{3} × 1 + a_{4} × 3 + a_{5} × 2)\)
\(= (-2a_{5} – 3a_{4} – a_{3} + 2a_{2} + 3a_{1} + a_{0})\)

\(N = (100002a_{5} – 2a_{5}) + (10003a_{4} – 3a_{4}) + (1001a_{3} – a_{3}) + (98a_{2} + 2a_{2})\) \(+ (7a_{1} + 3a_{1}) + a_{0}\)

\(N = 7(14286a_{5} + 1429a_{4} + 143a_{3} + 14a_{2} + a_{1}) + (-2a_{5} – 3a_{4} – a_{3} + 2a_{2}\) \(+ 3a_{1} + a_{0})\)

\(N = 7(14286a_{5} + 1429a_{4} + 143a_{3} + 14a_{2} +a_{1}) + 7k\)

So, if M is a multiple of 7, then so is N.
This can easily be generalized to any number of digits.

Note: Rule – 1-3-2 can be used for any number with 2 or more digits. It can help us to find the divisibility of any number by 7 easily and quickly as well.

Comparison

Explorations such as this help teachers plan lessons in which students develop capacities for problem-solving, logical reasoning, and computational thinking. Students become comfortable in working with abstractions and other core techniques of Mathematics and Computational Thinking, such as the mathematical modelling of phenomena and the development of algorithms to solve problems. (NCF-SE 2023).

If the teaching of divisibility rules stops at practising number skills, then we are severely limiting the potential of such a rich topic. Asking why the rule works, trying to generalise it, comparing different rules and then trying to make their own rules will not just develop mathematical minds but also impart an understanding of the joy and beauty of the subject.