Fun with Mathematics

Let n be the number n =(9999….9)²₂₀₀₆ – (666666….6)²₂₀₀₆ Find the remainder when n is divided by (a)11 (b)8 and (c)7.

 

As there are even number of  9s and 6s in the base numbers(2006 9s and 2006 6s) ,each factor will be divisible by 11. So the remainder will be zero when n is divided by 11.

Instead if there are odd number of 9s and 6s in the base number, It can be expressed as (y+9)² – (y’+6)² where y and y’ are divisible by 11. Y²+18y+81 –(y’²+12y’+36) = the remainder is determined by 81-36=45 44 is divisible by 11, so the remainder is 1. This is true for all odd number of 9s and 6s in the base numbers.

It is noted that when n is divided by 3 (like 11) ,both the factors are divisible by 3 and hence the remainder will be zero.

 The number n will always end with  45

n-5 is divisible by 8 (5x8 = 40) hence the remainder is 5 whatever may be the number of 9s and 6s.

Since n ends with 5 , the remainder will be zero when it is divided by 5.

9²- 6² = 45 ; the remainder when divided by 7  is 3

99² – 66² = 5445 the remainder when divided by 7 is 6 ;(y+1)² – (y’ +3)² ;y² +2y +1 – y’² -6y’ -9; 14+1 -9 = 6

164 x 33 = (y+4)(y’+5) ,the remainder is determined from 20(=4x5) , the remainder 20-14 = 6

999² – 666² = 554445,the remainder when divided by 7 is 3

9999²- 6666² = 55544445, the remainder is 5

99999²- 66666² = 5555444445, the remainder is 5
999999²- 666666² = 555554444445, the remainder is 0

It is observed that the remainder of each square varies cyclically, in the case of all 9s,when divided by 7

No.of 9s      remainder        no.of 6s   remainder

      1                   2                        1                 6

      2                   1                        2                 3

     3                    5                        3                  1

      4                   3                        4                   2

      5                   4                         5                  5

      6                   0                         6                   0              

 Since there are 2006 9s and 2006 6 s, when divided by 6  it gives a remainder 2.

Hence (999999….9)₂₀₀₆ is divided by 7 ,it will give a remainder 1

(666666…6)₂₀₀₆ is divided by 7 , it will give a remainder 3. The resultant remainder can be determined from (y+1)² + (y’+3)² All terms containing y and y’ will be divisible by 7 , the remaining term is 1+9 = 10. So the ultimate remainder is 3