December 9, 2022

# Divisibility rules or Divisibility tests

A number is said to be divisible by another number if the remainder is zero. Divisibility rules or divisibility tests are a set of general rules that are often used to determine whether or not a number is completely divisible by another number.  In this article,  some  divisibility tests are given with many examples .

Contents

# Divisibility by  powers of 2 and 5

Let $\small&space;n$

any integer , then  $\small&space;n$ is divisible by    $\small&space;2^{j}$ if the integer made up of  the last  $\small&space;j$  digits of  $\small&space;n$ are divisible by  $\small&space;2^{j}$ .

Suppose we have   to determine whether an integer is divisible by 2, we only need to examine its last digit for divisibility by 2.  To  determine whether $\small&space;n$ is divisible by 4 which is equal to $\small&space;2^{2}$   , we only need to check the integer made up of the last two digits of  $\small&space;n$  for divisibility by 4. Similarly to  determine whether $\small&space;n$  is divisible by 8 which is equal to  $\small&space;2^{3}$   , we only need to check the integer made up of the last three digits of $\small&space;n$   for divisibility by 8 and so on.

Example. Let$\small&space;n=32688048$.  we see that

(i)  $\small&space;n$  is divisible by 2 since its last digit  8  is divisible by 2.

(ii)  $\small&space;n$  is divisible by  $\small&space;2^{2}$ since its last 2 digit  48  is divisible by 4.

(iii) $\small&space;n$ is  divisible by $\small&space;2^{3}$ since its last 3  digit 048  is divisible by 8.

(iv) $\small&space;n$   is divisible by  $\small&space;2^{4}$ since its last 4 digit 8048 is divisible by 16.

(v)  $\small&space;n$  is  not divisible  $\small&space;2^{5}$ by  since  its last 5 digit 88048 is not  divisible by 32.

Divisibility tests for powers of 5 are same to those for powers of 2. We only need to check the integer made up of the last   $\small&space;j$   digits of  $\small&space;n$ to determine whether $\small&space;n$  is divisible by  $\small&space;5^{j}$ .

Example.   Let n=  15535375. we see that

(i)  $\small&space;n$ is divisible by 5 since its last digit  5  is divisible by 5

(ii) $\small&space;n$  is divisible by $\small&space;5^{2}$  since its last 2 digit  75  is divisible by 25.

(iii) $\small&space;n$ is   divisible by $\small&space;5^{3}$   since its last 3  digit 375  is divisible by 125.

(iv) $\small&space;n$ is not divisible by $\small&space;5^{4}$ since its last 4 digits 5375 is not divisible by 625.

## Divisibility test of 3 and 9

To see whether is divisible by 3, or by 9, we  only need to check whether the sum of the digits of is divisible by 3, or by 9.

Example : Let $\small&space;n$ = 4127835. Then, the sum of the digits of $\small&space;n$ is  4+ 1 +2+ 7 + 8 + 3 + 5=30. Since 30 is divisible by 3 then the given number is also divisible by 3.  But 30 is not divisible by 9 , so the given number is not divisible by 9.

### Divisibility tests of 7

(i)  Double the last digit and subtract it from a number made by the other digits. The result must be divisible by 7.

Example : Check whether 784 is divisible by 7.

Solution :In the given number 784, twice the digit in one’s place is =  2 ⋅ 4=  8

The number formed by the digits except the digit in one’s place is=  78

The difference between twice the digit in one’s place and the number formed by the other digits is =  78 – 8=  70. Since 70 is divisible by 7. So, the given number 784 is  divisible by 7.

(ii) A 12-year old Nigerian boy, Chika Ofili, made history this year after his new discovery in the field of mathematics. He was awarded at the  TruLittle Hero Awards for discovering the new divisibility test of 7, popularly called as Chika’s Test.

Chika’s Test

It is similar to the previously foregoing rule, but it is simpler and faster than it.

Step 1: Separate the last digit of the number.

Step 2: Multiply the last digit by 5 and add it to the remaining number.

Step 3: Repeat the steps unless you get a number within 0-70.

Step 4: If the result is divisible by 7, the number you started with is also divisible by 7.

Example: Let’s check divisibility of 23576  by 7

Sol.  2357+6×5= 2387

238+7×5=273

27+3×5=42.

Since 42 is divisible by 7, hence the given number is divisible by 7.

#### Divisibility tests of 11

(i) Using alternating sums: The given integer  is divisible by 11 , if and only if  the integer formed by alternately adding  and subtracting the digits, is divisible by 11.

Example : . We see that 723160823 is divisible by 11, since alternately adding and subtracting its digits yields        3-2+8-0+6-1+3-2+7=22 which is divisible by 11. On the other hand, 33678924 is not divisible by 11.
since 4-2+9-8+7-6+3 – 3 =4 is not divisible by 11.

(ii) Using pairs of digits

step1. Write down the number.

step 2. Divide the digits in pairs from right to left. (The last digit on the left might be alone)

step 3. Add the numbers together .

step 4. Check whether the answer is divisible by 11.   If it is, then the original number is divisible by 11 as well

Example.   Let’s check  $\small&space;n$= 17957 is divisible by 11 or not.

Divide the digits in pairs from right to left

 1 79 57

on adding  the numbers together we get 1+79+57= 137. Now check whether the answer is divisible by 11.

On repeating the same steps to check whether 137 is divisible by 11 or not , we get  1+37=38 . Since 38 is not divisible by 11, then 137 is as well. Since 137 is not divisible by 11,  our  original number 17957 is also  not  divisible by 11.

##### Divisibility by 7,11 and 13

Step 1. Write the given number.

Step 2 . Make blocks of three digits from right to left.

Step 3. Now find alternating sum and difference of blocks of three digits.

step 4 : Check whether the  answer  is divisible by 7,11 and 13. If it is, then the original number is divisible by 7, 11 and 13 as well

For example : (i)  Let n = 59358208. Since the alternating sum and difference of the integers formed from blocks of three digits, 208 – 358 + 59 = -91, is divisible by 7 and 13, but not by 11. We see that  n is divisible by 7 and 13 but n is not divisible by 11.

(ii)Let n= 525120596. Since the alternating sum and difference of the integers formed from blocks of three digits  is 596 – 120 + 525 = 1001.  Since 1001 is divisible  by 7, 11 and 13. We see that  n is also divisible by  7 , 11 and 13.