Remainder of power numbers on division

Remainder of power numbers on division

The remainder of a number n to the power of p when divided by a divisor d is cyclically varying not only for p but also for n . For example when d = 7 ,the cyclic variation of the remainder is same for n^p  and

(n+ 7x)^p

Remainder of n^p when divided by 7

n/p         0              1              2              3              4              5              6              7              8              9              10

1              1              1              1              1              1              1              1              1              1              1              1

2              1              2              4              1              2              4              1              2              4              1              2             

3              1              3              2              6              4              5              1              3              2              6              4

4              1              4              2              1              4              2              1              4              2              1              4

5              1              5              4              6              2              3              1              5              4              6              2

6              1              6              1              6              1              6              1              6              1              6              1

7              1              0              0              0              0              0              0              0              0              0              0

8              1              1              1              1              1              1              1              1              1              1              1             

9              1              2              4              1              2              4              1              2              4              1              2

10           1              3              2              6              4              5              1              3              2              6              4

It is found that the remainder of (1+7x)^p when divided by 7 is 1,for all values of x and p..The remainder of  n⁶ when divided by 7  is 1 for all values of n except n = 7x. It is 0 for n = 7x

power numbers and its divisibility

Power numbers and its divisibility

When a seventh power of a number ‘a’ is divided by 7, the remainder will be equal to that number itself when a < 7 and (a-7n) when a > 6 where n is the  whole number of 7’s in ‘a’ . e.g.,

1⁷ = 1 = 7(0) + 1

2⁷ = 128 = 7(18) + 2

3⁷ = 2187 = 7(312) + 3

4⁷ = 16384 = 7(2340) + 4

5⁷ = 78125 = 7(11160) + 5

6⁷ = 279936 = 7(39990) + 6

When a = 7 or > 7, a= (7n+d)

a⁷ = (7n+d)⁷ and the remainder is determined by d⁷ / 7 and it is equal to d itself.

7⁷ = 823543 = 7 (117648) + 7

                      = 7 (117649) + 0

8⁷ = 2097152 = 7 (299593) + 8

                        = 7 (299594) + 1

 When a fourth power of a number ‘a’ is divided by (a+1) ,the remainder is 1 for all values  of a.

a=1; a⁴ = 1 = 2(0) + 1

a=2; a⁴ = 16 = 3(5) + 1

a=3; a⁴ = 81 = 4(20) + 1

a=4; a⁴ = 256= 5(51) + 1

a=5; a⁴ = 625 = 6(104) + 1

a=6; a⁴ = 1296 = 7(185) + 1

a=7; a⁴ = 2401 = 8(300) + 1

a=8; a⁴ = 4096 = 9(455) +1

The quotient is equal to a³ – a² + a - 1

divisibility of cube numbers (cont.,)

Divisibility of cube numbers (cont.,)

When a³ is divided by (a+4),,the remainder will be 4(a-12),that is for all ‘a’ which are in multiples of 12 ,the remainder will be zero.

The general form correlating a³, the divisor (a+4),the quotient and the remainder is

  a³ = [8+ (a-2)² ] (a+4) + 4(a-12)

If the  divisor is (a+5),  it becomes,

a³ = [14 + (a-2)(a-3)] (a+5) + 5(a-20)

For the divisor (a+6),

a³ = [21 + (a-3)² ] (a+6) + 6(a-30)

for (a+7),

a³ = [ 30 + (a-3)(a-4)](a+7) + 7(a-42)

divisibility of cube numbers (cont.,)

Divisibility of cube numbers (cont.,)

When a³ is divided by 4 ,for all even numbers the remainder is zero where as for odd numbers it is 1 for a in the form of (4n+1) and 3 for (4n-1).

When a³ is divided by (a+4)

1= 5(9) -44

8= 6(8) – 40

27 = 7(9) – 36

64= 8(12) – 32

125 = 9(17) – 28

216 = 10(24) – 24

343= 11(33) – 20

512 = 12(44) – 16

729 = 13(57)- 12

1000 = 14(72) – 8

 In  general it takes a form

a³ = [8+(a-2)²](a+4) + 4(a-12)

divisibility of cube numbers (cont.,)

Divisibility of cube numbers (cont.,)

When a³  is divided by 3, the remainder will not only be same as the remainder for a but also it is periodically varying as 1,2,0.

When a³ is divided by (a+3)

a=1, 4(4) – 15

a=2,5(4) – 12

a=3, 6(6) - 9

a=4,7(10) - 6

a=5,8(16) - 3

a=6,9(24) + 0

a=7,10(34) + 3

a=8,11(46) + 6

a=9,12(60) + 9

                                                                            a-2

In general ,the quotient is in the form [4 +  Σ 2n ] and the remainder is 3(a-6)

                                                                                     0        

a-2

 Σ 2n  = (a-1)(a-2)

 0

It shows that a³ = [4+(a-1)(a-2)](a+3) - 3(a-6)

divisibility of cubes

Divisibility of cubes

We know that a³ – a is equal to a product of three successive numbers (a-1),a,and (a+1).

a³ – a = (a-1)a(a+1). Since the product of any three successive numbers is invariably divisible by 6,

(a-1)a(a+1) = 6n

i.e., a³ – a = 6n  or a³ = a+6n or a² = 1 + [6n/a] i.e., 6n is divisible by a

It is noted that there is some kin d of regularity in the quotient and the remainder when a³ is divided by (a+1),(a+2),(a+3)…..

When a cube a³ is divided by (a+1) or (a-1) ,the quotient will be a(a-1) or a(a+1) with remainder a.

When a³ is divided by (a+2)

a=1 ,2(3) – 5

a=2, 3(4) – 4

a=3, 6(5) -3

a=4,11(6) – 2

a=5,18(7) – 1

a=6,27(8) + 0

a=7, 38(9) + 1

a=8,  51(10) +2

a=9, 66(11) +3

In general, it is in the form,

               a-2

a³ = [ 2+ Σ (2n+1)](a+2) + (a-6)

           0

But               n-2

Σ (2n+1)] is a square number.It is equal to square of a number equal tp the

0

  number of odd numbers added together.i.e, (n-1)² .

a³ = [2+(a-1)²](a+2) + (a-6)

It shows that when a³ is divded by (a+2) , the quotient will be [2+(a-1)²] and the remainder will be (a-6)