**Divisibility Rule for number 7**

Divisibility rules involve finding out without actually dividing whether a given whole number is a factor of another whole number.

A number is divisible by 7 if:

- The sum of the last digit and 3 times the rest is a multiple of 7
- The sum of 3 times the last digit and twice the rest is a multiple of 7
- the sum of the number formed by the last two digits and twice the remaining number is divisible by 7
- the SUM of the weighted digits in the odd and even groups is divisible by 7 ( where the unit digit, ten digit and hundred in the odd groups are assigned positive 1, 3, and 2 and the even groups are assigned negative values of 1, 3, 2 respectively. This rule uses modulo 7 to derive these weighted values from powers of 10)

Note: The number to be tested will be grouped in threes starting from right to left and assigned the weight (2, 3, 1). The last group may not be up to three

**Rule 1:**

**A number is divisible by 7 if the sum of the last digit and 3 times the rest is a multiple of 7**

Proof

Let N be a number divisible by 7

Consider N = 10t + u and N_{4} = 3t + u

7 divides N_{4} if and only if 7 divides 8N_{4}

8N_{4} = 8( 3t + u)

= 24t + 8u

= 10t + u +14t +7u

= 10t + u + 7(2t + u)

Since 7 divides 7(2t + u) and 7 divides N, therefore 7 divides **8N _{4}**

**Example 1: **Test that 238 is divisible by 7

The number becomes 23 and 8

Multiply 23 by3 and add it to 8 => 69 + 8

** = 77**

Since 77 is divisible by 7, then 238 is divisible by 7

###### Rule 2

**A number is divisible by 7 if the sum of 3 times the last digit and twice the rest is a multiple of 7 **

**Proof:**

**Let N = 10r + u **that is a number divisible by 7

**where u is the unit digit and r is the remaining number**

Consider N_{3 }= 2r + 3u

7 divides N_{3} if and only if 7 divides 12 N_{3}

12 N_{3 } =12(2r + 3u)

** = **24r + 36u

= 10r + u + 14r + 35u

= 10r + u + 7(2r + 5u)

= N + 7(2r + 5u)

Since 7 divides 7(2r + 5u) and 7 divides N, therefore 7 divides **12N _{3}**

**Example 2:** Test that 203 is divisible

Separate the number into unit digit and rest, that is 3 and 20

Multiply 2 by 20 and 3 by 3 . This gives 40 + 9= 49

Since 49 is divisible by 7, 203 is divisible by 7

**Example 3:** Test that 315 is divisible by 7

** The numbers becomes 31 and 5**

**Therefore, ( 2 X 31) + (3 X 5)**

** =62 + 15**

** = 77**

Since 77 is divisible by 7, then 315 is divisible by 7

###### Rule 3

**A number is divisible by 7 if the sum of the number formed by the last two digits (ten and unit digits) and twice the rest is divisible by 7**

Let N = 100H +10T + U (that is a number divisible by 7)

Where U is the unit digit, T is the ten digit and H is the remaining digits

Consider Ni = 2H + (10T +U)

7 divides Ni if 7 divides 50Ni

50Ni = 50(2H + (10T + U))

=100H + 500T + 50U

= (100H +10T + U) + 490T + 49U

= N + 49(10T + U)

Since 7 divides N and divides 49(10T + U), 7 divides 50Ni

**Example 4: **Use Rule 3 to test that 1043 is divisible by 7, that is adding the last two digits to twice the rest

the last two digits form 43

twice the rest number is 2 x 10 = 20

the sum gives 63 which is a number divisible by 7

Since 63 is divisible by 7 therefore 1043 is a multiple of 7

###### Rule 4

**A number is divisible by 7 if the SUM of the weighted digits in the odd and even groups is divisible by 7 ( where the unit digit, ten digit and hundred in the odd groups are assigned positive 1, 3, and 2 and the even groups are assigned negative values of 1, 3, 2 respectively) **

**Note: the number to be tested will group in threes starting from right to left and assigned the weight (2, 3, 1)**

Let N = a_{0 } + 10a_{1 } + 100a_{2}+ 1000a_{3 }+ 10000 a_{4 } + 100000a_{5 }+… a number divisible by 7

Where a_{0} is the unit digit,

a_{1} is the ten digit,

a_{2} is the hundred digit,

a_{3} is the thousand digit, _{ } _{ }

a_{4} is the ten-thousand digit

Consider N_{2 } = a_{0} + 3 a_{1} + 2 a_{2} – a_{3} – 3 a_{4}– 2 a_{5 } + ….

7 divides _{ }N_{2} if and only if 7 divides 8 N_{2}

8 N_{2} = 8(a_{0} + 3 a_{1} + 2 a_{2} – a_{3} – 3 a_{4}– 2 a_{5 } + ….)

= 8a_{0} + 24 a_{1} + 16 a_{2} – 8a_{3} – 24a_{4 }– 16a_{5 } + ….

Manipulating each of the values above e. g 8a is the same a + 7a and 24a = 10a + 14a

= a_{0 } + 10a_{1 } + 100a_{2}+ 1000a_{3 }+ 10000 a_{4 } + 100000a_{5 }+

7a_{0} + 14a_{1} – 84a_{2} – 1008 a_{3} – 10024a_{4}– 100016a_{5 } + ….

= N + 7(a_{0 } + 2a_{1 } + 12a_{2}+ 144a_{3 }+ 1432a_{4 } + 14288a_{5 }+… )

Since 7 divides 7(a_{0 } + 2a_{1 } + 12a_{2}+ 144a_{3 }+ 1432a_{4 } + 14288a_{5 }+…) and 7 divides N, then 7 divides 8 N_{2.}

###### Examples

**1**. Test if 7 is a factor of 100125

Using the rule, group the numbers in threes starting from the unit digit

100 | 125 where 125 forms the odd group and 100 the even group

Assigning the weighted values for the odd group = 8 X1 + 3 X2 +1 X2 = 8 + 6+ 2=16

Assigning the weighted values for the even group = 0 X (-1) + 0 X (-3) + 1X (-2) = -2

The sum of the two groups is 14 which is a number divisible by 7

Since 14 is divisible by 7 therefore 100125 is divisible by 7

**2.** Test that 58967041 is divisible by 7

these are the weighted values 3|1|-2 |-3 |-1 | 2 | 3 | 1

Number to test 5|8 |9 | 6 | 7 | 0 | 4| 1

Multiplying the numbers by the weights= 15 + 8 -18 -18 -7 + 0 +12 + 1= -7

Since – 7 is divisible by 7 therefore 58967041 is divisible by 7

_{Note: The Rule 3 and Rule 4 are in Wikipedia but the proofs are my work}

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