Rules for Divisibility – Part One

The tantalizing and compelling pursuit of mathematical problems offers mental absorption, peace of mind amid endless challenges, repose in activity, battle without conflict, refuge from the goading urgency of contingent happenings, and the sort of beauty changeless mountains present to senses tried by the present day kaleidoscope of events.

Morris Kline

With the passage of time, we felt the need for tricks and techniques to solve maths problems faster and easily without lengthy calculation. In this blogpost, we will discuss the rules of divisibility for the numbers 1 to 11 with examples. this will help to to check whether a number is divisible by another number from 1 to 11 without the actual method of division. To begin with, consider the following definition.

Definition. An integer a is said to be divisible by a number b (≠ 0) if there is an integer c such that a = bc. In mathematics, we denote this fact by b | a, read as b divides a.

Since every number is divisible by 1, divisibility rule for 1 doesn’t have any condition. While dividing by 1, we will get back the same number.

A number is divisible by 2 if it ends with 0, 2, 4, 6 or 8. That is, the digit in the unit place are either 0 or 2 or 4 or 6 or 8.

Example Consider the number 582. Since that last digit is 2, the number is divisible by 2. But 5975 is not divisible by 2, since the last digit 5 is not divisible by 2.

A number is divisible by 3 if the sum of the digits of the number is divisible by 3.

Example Consider the number 580. Sum of the digits of this number = 5 + 8 + 0 = 13, which is not divisible by 3. Hence 580 in not divisible by 3. But 582 is divisible by 3, since the sum of the digits 5 + 8 + 2 = 15, which is divisible by 3.

Let’s prove it for a four-digit number. Consider a four-digit number abcd, which can be written as

1000a + 100b + 10c + d

= (999a + 99b + 9c) + (a + b + c + d)

= 3(333a + 33b + 3c) + (a + b + c + d),

applying associativity and commutativity of addition. The first term 3(333a + 33b + 3c) is always divisible by 3. The second addend (a + b + c + d) is the sum of the digits. Therefore, the number is divisible by 3 if and only if the sum of its digits is divisible by 3.

Even though this proof refers to a four-digit number, it gives a general idea how the proof can be extended to a number with n-digits. The strategy used in this proof is representing a number as a sum of two addends. The divisibility of one component is obvious. The divisibility of the second component determines the divisibility of the number. A similar strategy will be applied in the following proofs.

Alternatively, a number is divisible by 3 if −2 times the last digit of the number added to the rest of the number is divisible by 3.

Let’s prove it for a three-digit number. Consider a three-digit number abc, which can be written as 100a + 10b + c.

The two-digit number is ab, obtained after deleting the last digit c. Hence, −2 times the last digit of the number added to the rest of the (two-digit) number = 10a + b −2c. We have

10(10a + b −2c)

= 100a + 10b − 20c

= 100a + 10b + c − 21c.

Therefore, abc (= 100a + 10b + c) = 10(10a + b −2c) + 21c.

It follows that, abc is divisible by 3 if 10a + b −2c is divisible by 3.

Example Consider the number 342. The last digit is 2 and the two-digit number after removing the last digit is 34. Now, −2 times the last digit of the number added to the rest of the (two-digit) number = 34 – 4 = 30, which is divisible by 3. Hence the original number is divisible by 3. But, 530 is not divisible by 3 because, 53 – 0 = 53 is not divisible by 3.

A number is divisible by 4, if it’s last two digits, when read as a two-digit number, is divisible by 4. If the number is of one digit, one can apply the rule directly.  A fast way to divide by 4 is to divide the original number by 2, and then by 2 again.

Example Consider the number 484. The last two digits of the number is 84, which is divisible by 4. Hence 484 in divisible by 4. But 97 is not divisible by 4, since this two-digit number is not divisible by 4.

A number is divisible by 5 if it ends with 0 or 5.

Example Consider the number 484. The last two digit of the number is 4. Hence 484 is not divisible by 5. But 1365 is divisible by 5, since the last digit of the number is 5.

Let’s prove it for a four-digit number. Consider a four-digit number abcd, which can be written as

1000a + 100b + 10c + d = 5(200a + 20b + 2c) + d.

It is evident from the above representation that, abcd is divisible by 5 if d is either 0 or 5.

A number is divisible by 6 if it is divisible by 2 and 3. So, if a number is even and if the total of its digits can be divided by 3, then the original number is divisible by 6.

Example Consider the number 744. This number is divisible by 2. Also, sum of the digits = 7 + 4 + 4 = 15, which is divisible by 3. Hence the number is divisible by 3. But the number 946 is not divisible by 6, as it is not divisible by 3.

The rule is not so straightforward. To check if a number is divisible by 7, take the last digit off the number, double that digit and subtract the doubled number from the remaining number. If the result is divisible by 7, it may be negative, then the number is divisible by seven. This may need to be repeated several times.

Example Consider the number 896. The last digit is 6 and double of this is 12. The two-digit number after removing the last digit is 89. Now 89 – 12 = 77. Since, 77 is divisible by 7 so the original number, 896 is also divisible by 7. Similarly, 895 is not divisible by 7.

Let’s prove it for a three-digit number. Consider a three-digit number abc, which can be written as 100a + 10b + c.

The two-digit number is ab, obtained after deleting the last digit c. Hence, −2 times the last digit of the number added to the rest of the (two-digit) number = 10a + b −2c. We have

10(10a + b −2c)

= 100a + 10b − 20c

= 100a + 10b + c − 21c.

Therefore, abc (= 100a + 10b + c) = 10(10a + b −2c) + 21c.

It follows that, abc is divisible by 7 if 10a + b −2c is divisible by 7.

Alternatively, write a number N as N = am(1000)m + am − 1(1000)m − 1 + ··· + a1(1000) + a0, where ak are integers and 0 ≤ ak ≤ 999. Let T = a0a1 + a2a3 + ···. Then N is divisible by 7 if T is divisible by 7.

Example Consider the number N = 23146123. We can write it as N = 23(1000)2 + 146(1000) + 123. Now, N is divisible by 7 because the sum 123 – 146 + 23 = 0 is divisible by 7.

A number is divisible by 8, if the number formed by the last three digits of the number is divisible by 8. A fast way to divide by 8 is to divide the original number by 2, then by 2, and finally by 2 again.

Example The number 8024 is divisible by 8 because 024 (= 24) is divisible by 8. But the number 3694806 is NOT divisible by 8 because 806 is not divisible by 8.

A number is divisible by 9 if the sum of the digits of the number is divisible by 9.

Example The number 3456 is divisible by 9 because the sum of the digits of the number = 3 + 4 + 5 + 6 = 18 is divisible by 9. But the number 1234 is not divisible by 9 because the sum of the digits of the number = 1 + 2 + 3 + 4 = 10 is not divisible by 9.

A number is divisible by 10 if it ends with 0. In other words, a number is divisible by 10 if it is even and it is divisible by 5.

Example The number 369480 is divisible by 10 because the last digit is 0.

Add the digits in the odd positions of the number; add the digits in the even positions in the number. Subtract the two sums. If the result is 0 or is divisible by 11, then the original number is divisible by 11.

Example Consider the number 1234. First, odd place digits are 4 and 2. Sum of the digits = 4 + 2 = 6. Even place digits are 3 and 1. Sum of the digits = 3 + 1 = 4. Difference of the sums = 6 – 4 = 2, which is not divisible by 11. Therefore, 1234 is not divisible by 11. Consider the number 1234567. Sum of its digits in the odd places = 7 + 5 + 3 + 1 = 16. Sum of its digits in even places = 6 + 4 + 2 = 12. Their difference = 16 – 12 = 4, which is not divisible by 11. So 1234567 is not divisible by 11.

Alternatively, a number is divisible by 11 if −1 times the last digit of the number added to the rest of the number is divisible by 11. This may need to be repeated several times.

Example Consider the number 9647. The last digit is 7 and the three-digit number after removing the last digit is 964. Now, −1 times the last digit of the number added to the rest of the (three-digit) number = 964 – 7 = 957. Repeating the procedure again, we get 95 – 7 = 88, which is divisible by 11. Hence 9647 is divisible by 11.

Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call Math1089 – Mathematics for All!“.

2 comments

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