Divisibility Rule for 7 with proofs

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 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 divides 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

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 =  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.

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

Divisibility Rules For numbers 10 – 19

Divisibility Rules For numbers 10 – 19

The rules for numbers greater than 10 especially numbers between 10 and 20 work in this way – Given a number y = 10T + U, where U is the unit digit number and T is the number formed by the rest digits.

A number y is divisible by N (where N is a number greater than 10) if (N-10)T – U is divisible by N. That is, find the difference between the number N and 10

Note: This is true of all numbers greater than 10 but is not efficient in some cases

The rule for 11 using the above:

A number y is divisible by 11 if and only if (11 – 10)T – U is divisible by 11

The difference between 11 and 10 is 1

Example 1: Test whether 132 is divisible by 11,

Multiply 13 by 1 and subtract 2 from the product. That is (1×13) – 2 = 13 -2 = 11,  a number divisible by 11.

Example 2: Test whether 1023 is divisible by 11,

Multiply 102 by 1 and subtract 3 from the product. That is (1×102) – 3 =  102 – 3 = 99 a number divisible by 11.

The rule for 12

Old: A number is divisible by 12 if and only if the number is divisible by 3 and 4.

Using the above: A number y is divisible by 12 if and only if (12 -10)T – U (that is 2T – U ) is divisible by 12

The difference between 12 and 10 is 2

Example 1: Test whether 156 is divisible by 12,

Multiply 15 by 2 and subtract 6 from the product. That is (2 x 15) – 6 = 30 – 6 = 24, gives a number divisible by 12.

Example 2: Test whether 576 is divisible by 12,

Multiply 57 by 2 and subtract 6 from the product. That is (2 x 57) – 6 = 114 – 6 = 108, a number divisible by 12.

The rule for 13

Using the above: A number y is divisible by 13 if and only if (13 -10)T – U (that is 3T – U ) is divisible by 13.

The difference between 13 and 10 is 3

Example 1: Test whether 169 is divisible by 13,

Multiply 16 by 3 and subtract 9 from the product. That is (3 x 16) – 9 = 48 – 9 = 39, a number divisible by 13.

Example 2: Test whether 312 is divisible by 13,

Multiply 31 by 3 and subtract 2 from the product. That is (3 x 31 ) – 2 = 93 – 2 = 91, a number divisible by 13

Rules for 14

(a) A number is divisible by 14 if it is divisible by 2 and 7

(b)  A number y is divisible by 14 if and only if (14 -10)T – U (that is 4T – U ) is divisible by 14.

(c)  A number is divisible by 14 if 6T + 2U is divisible by 14

Using rule (b)

The difference between 14 and 10 is 4

Example 1: Test whether 98 is divisible by 14,

Multiply 9 by 4 and subtract 8 from the product. That is (4 x 9) – 8 = 36 – 8 = 28, a number divisible by 14.

Using rule (c)

Example : Test whether 182 is divisible by 14,

182: (6 x 18 ) +( 2 x 2) = 108 + 4 = 112,

112: (6 x 11) + 2 x 2 = 70 = 14 x 5,  a number divisible by 14.

The rule for 15

(a): A number is divisible by 15 if and only if the number is divisible by 3 and 5.

(b) : A number y is divisible by 15 if and only if (15 -10)T – U (that is 5T – U ) is divisible by 15.

(c) : A number y is divisible by 15 if and only if  5T + 2U ) is divisible by 15.

Using rule (b)

The difference between 15 and 10 is 5

Example 1: Test whether 135 is divisible by 15,

135:  (5 x 13) – 5 = 65 – 5 = 60 = 15 x 4, a number divisible by 15.

Using rule (c)

Example : Test whether 225 is divisible by 15,

225:  is (5 x 22 )  + (2 x 5) = 110 + 10 = 120 = 15 x 8, a number divisible by 15.

The rule for 16

(a) A number y is divisible by 16 if and only if (16 -10)T – U (that is 6T – U ) is divisible by 16.

(b) A number is divisible by 16 if 4T + 2U is divisible by

The difference between 16 and 10 is 6

Example 1: Test whether 128 is divisible by 16,

Multiply 12 by 6 and subtract 8 from the product. That is (12 x 6) – 8 = 72 – 8 = 64, a number divisible by 16.

Example 2: Test whether 256 is divisible by 16,

Multiply 25 by 6 and subtract 6 from the product. That is (6 x 25 ) –  6 = 150 – 6 = 144, a number divisible by 16.

The rule for 17

(a): A number y is divisible by 17 if and only if (17 -10)T – U (that is 7T – U ) is divisible by 17.

(b) A number is divisible by 17 if 3T + 2U is divisible by 17

Using rule (a)

The difference between 17 and 10 is 7

Example : Test whether 68 is divisible by 17,

68:  (7 x 6) – 8 = 42 – 8 = 34 = 17 x 2, a number divisible by 17.

Using rule (b)

Example 2: Test whether 204 is divisible by 17,

204: (7 x 20) – 4 = 140 – 4 = 136, a number divisible by 17.

If one is not sure that 136 is divisible by 17, this number can be tested recursively. That is, (7 x 13) – 6 = 91 – 6 = 85= 17 x 5, a number divisible by 17.

The rule for 18

(a): A number is divisible by 18 if and only if the number is even and divisible by 9.

(b): A number y is divisible by 18 if and only if (18 -10)T – U (that is 8T – U ) is divisible by 18.

(c): A number is divisible by 18 if 2T + 2U (that is, 2(T + U) is divisible by 18

Using rule (b)

The difference between 18 and 10 is 8

Example : Test whether 72 is divisible by 18,

72:  (8 x 7) – 2 = 56 – 2 = 54 = 18 x 3, a number divisible by 18.

Using rule (c) that is testing for 2 and (T + U)

Example : Test whether 126 is divisible by 18,

Since 126 is even and 126: 12 + 6 = 18, 126 is divisible by 18.

The rule for 19

Using the above: A number y is divisible by 19 if and only if (19 -10)T – U (that is 9T – U ) is divisible by 19.

The difference between 19 and 10 is 9

Example 1: Test whether 76 is divisible by 19,

Multiply 7 by 9 and subtract 6 from the product. That is (9 x 7) – 6 = 63 – 6 = 57, a number divisible by 19.

Example 2: Test whether 209 is divisible by 19,

Multiply 20 by 9 and subtract 9 from the product. That is (9 x 20) – 9 = 180 – 9 = 171, a number divisible by 19.

If one is not sure that 171 is divisible by 19, this number can be tested recursively. That is, (9 x 17) – 1 = 153 – 1 = 152, which is a number divisible by 19.

A simpler Rule for 19:

A number y is divisible by 19 if and only if T + 2U is divisible by 19.

Example 1: Test whether 209 is divisible by 19,

Using T + 2U, 20 + (2 x 9) = 38, a number divisible by 19.

Example 2: Test whether 437 is divisible by 19,

Using T + 2U, 43 + (2 x 7) = 43 + 14 = 57, a number divisible by 19.