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.

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.

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.

Which of the following numbers are divisible by 7?

a) 133 b) 178 c) 847 d) 988 e) 10787

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

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