Q. 2

By Euclid Division Lemma show that the cube of any positive integral number is of the form 9q or 9q + 1 or 9q + 8, where q is an integral number.

Answer :

According to the Euclid’s Division Lemma,

Let a be any positive integer and b = 9


Then by Euclid’s Lemma we have,


a = 9q + r


Where r = 0, 1, 2, 3, 4, 5, 6, 7, 8 [since 0 ≤ r ≤ b] and here value of b is 9]


So all the possible forms of a are as follows:


9q


9q + 1


9q + 2


9q + 3


9q + 4


9q + 5


9q + 6


9q + 7


9q + 8


Now to find the cubes of these values we have the following expansion formula:


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(9q + b)3 = 729q3 + 243q2b + 27qb2 + b3


Now when we divide the above equation by 9 we get the quotient as


81q3 + 27q2b + 3qb2 and the remainder is b3.


So we have to consider the value of b3


So to put b = 0


We get 9m + 0 = 9m


So to put b = 1


We get 13 = 1 so we get = 9m + 1


So to put b = 2


We get 23 = 8 so we get = 9m + 8


So to put b = 3


We get 33 = 27 and it is divisible by 9, so we get = 9m


So to put b = 4


We get 43 = 64 and when divided by 9 we get 1 as remainder, so we get = 9m + 8


So to put b = 5


We get 53 = 125 and when divided by 9 we get 8 as remainder, so we get = 9m + 8


So to put b = 6


We get 63 = 216 and when divided by 9 we get 0 as remainder, so we get = 9m


So to put b = 7


We get 73 = 343 and when divided by 9 we get 1 as remainder, so we get = 9m + 1


So to put b = 8


We get 83 = 512 and when divided by 9 we get 8 as remainder, so we get = 9m + 8


So it is proved that all the values are in the form of 9m, 9m + 1 and 9m + 8.


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