# Divisibility Rule for 7 with proofs

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 N4 = 3t + u

7 divides N4 if and only if 7 divides 8N4

8N4 = 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 8N4

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 N3 = 2r + 3u

7 divides N3 if and only if 7 divides 12 N3

12 N=12(2r + 3u)

= 24r + 36u

= 10r + u + 14r + 35u

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

= N  + 7(2r + 5u)

Since 7(2r + 5u) and 7 divides N, therefore 7 divides 12N3

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

The difference between the sum of the weighted digits in the even  and odd 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 =  a0  + 10a + 100a2+ 1000a3 +  10000 a  + 100000a5 +…     a number divisible by 7

Where a0 is the unit digit,

a1 is the ten digit,

a2 is the hundred digit,

a3 is the thousand digit,

a4 is the ten-thousand digit

Consider N2    = a0 + 3 a1 + 2 a2  – a3 – 3 a4– 2 a5    + ….

7 divides    N2 if and only if 7 divides 8 N2

8 N2 = 8(a0 + 3 a1 + 2 a2  – a3 – 3 a4– 2 a5    + ….)

= 8a0 + 24 a1 + 16 a2  –  8a3 – 24a4 – 16a5    + ….

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

=  a0  + 10a + 100a2+ 1000a3 +  10000 a  + 100000a5   +

7a0 + 14a1 – 84a2 –  1008 a3 – 10024a4–  100016a5    + ….

= N + 7(a0  + 2a + 12a2+ 144a3 +  1432a  + 14288a5 +… )

Since 7 divides 7(a0  + 2a + 12a2+ 144a3 +  1432a  + 14288a5 +…) and 7 divides N, then 7 divides 8 N2.

Example 5 : 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

Example  6: 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 already in Wikipedia but the proofs are my work