# Divisibility Rule for 7 Examples and Questions

Examples and detailed solutions on the divisibility rule for 7 are presented. Questions and their solutions are also included.
More on the divisibility rules are included.

## Divisibilty Rule for 7

To test if a number is divisible by 7, we subtract twice the last digit (unit digit) of the number from the remaining number (with last digit removed). If the result is equal to 0 or is a multiple of 7, then the number is divisible by 7.
The one and two digit numbers that are divisible by 7 (or multiples of 7) are:      0 , 7 , 14 , 21 , 28 , 35 , 42 , 49 , 56 , 63 , 70 , 77 , 84 , 91 , 98.

## Examples

Example 1

Is 154 divisible by 7?
The last digit in the given number 15
4 (unit digit ) is 4 .
We now use the given number without the last digit which is 15.
Subtract twice the last digit 4 from 15:
15 - 2 (4) = 15 - 8 = 7
The result 7 is a multiple of 7 and therefore 154 is divisible by 7.
Checking using long division: 154 � 7 = 22 with remainder 0.

Example 2 - We may need to use the rule more than once
Is 903 divisible by 7?
Step 1
The last digit in the given number 90
3 (unit digit ) is 3 .
We now use the given number without the last digit which is 90.
Subtract twice the last digit 3 from 90:
90 - 2 (3) = 90 - 6 = 84
If it is still not easy to determine whenther the result is divisible by 7, we continue using the rule on the result 84 abtained in the last step.
Step 2
The last digit of 84 is 4 and the number without the last digit is 8
Subtract twice the last digit 4 from 8
8 - 2(4) = 8 - 8 = 0
Conclusion: the given number 903 is divisible by 7.
Checking using long division: 903 � 7 = 129 with remainder 0.

Example 3 - We may need to use the rule several times
Is 86415 divisible by 7?
Step 1
The last digit in the given number 8641
5 (unit digit ) is 5 .
We now use the given number without the last digit which is 8641.
Subtract twice the last digit 5 from 8641:
8641 - 2 (5) = 8641 - 10 = 8631
Step 2
We use the rule on the result 8631 obtained
The last digit of 8631 is 1 and the number without the last digit is 863
Subtract twice the last digit 1 from 863
863 - 2(1) = 863 - 2 = 861
Step 3
We use the rule on the result 861 obtained
The last digit of 861 is 1 and the number without the last digit is 86
Subtract twice the last digit 1 from 86
86 - 2(1) = 86 - 2 = 84
We have already seen above that 84 is divisible by 7.
Conclusion: the given number 86415 is divisible by 7.
Checking using long division: 86415 � 7 = 12345 with remainder 0.

## Questions

( with solutions )

Which of the following numbers are divisible by 7?
a) 133     b) 178     c) 847     d) 988     e) 10787

## Solutions to the Above Questions

a) step 1:     13 - 2(3) = 13 - 6 = 7
Conclusion: the last result 7 is a mutliplbe of 7 and therefore 133 is divisible by 7

b) step 1:     17 - 2(8) = 17 - 16 = 1 ,
Conclusion: the last result 1 is NOT a mutliplbe of 7 and therefore 178 is NOT divisible by 7

c) step 1:     84 - 2(7) = 84 - 14 = 70 ,
Conclusion: the last result 70 is a mutliplbe of 7 and therefore 847 is divisible by 7

d) step 1:     98 - 2(8) = 98 - 16 = 82 ,
step 2:     8 - 2(2) = 8 - 4 = 4
Conclusion: the last result 4 is NOT a mutliplbe of 7 and therefore 988 is NOT divisible by 7

e) step 1:     1078 - 2(7) = 1078 - 14 = 1064 ,
step 2:     106 - 2(4) = 106 - 8 = 98
Conclusion: the last result 98 is a mutliplbe of 7 and therefore 10787 is divisible by 7