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 154 (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 903 (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 86415 (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