Q. 43 X5.0( 1 Vote )

On division by 6, a2 cannot leave remainder (a N)
A. 1

B. 4

C. 5

D. 3

Answer :

From Euclid’s Division lemma, for any positive integer aN.


a = 6m + r where, 0<r<6 and m N


a = 6m, a = 6m + 1, a = 6m + 2, a = 6m + 3, a = 6m + 4 or a = 6m + 5


Now, a = 6m


Then a2 = 62m2


= 36m2 = 6 × (6m)2 + 0


Here, remainder is 0.


If a = 6m + 1


Then a2 = (6m + 1)2


= 36m2 + 12m + 1


= 6(6m2 + 2m) + 1


Here, remainder is 1.


If a = 6m + 2


Then a2 = (6m + 2)2


= 36m2 + 24m + 4


= 6(6m2 + 4m) + 4


Here, remainder is 4.


If a = 6m + 3


Then a2 = (6m + 3)2


= 36m2 + 32m + 9


= 6(6m2 + 6m + 1) + 3


Here, remainder is 3.


If a = 6m + 4


Then a2 = (6m + 4)2


= 36m2 + 48m + 16


= 6(6m2 + 8m + 2) + 4


Here, remainder is 4.


If a = 6m + 5


Then a2 = (6m + 5)2


= 36m2 + 60m + 25


= 6(6m2 + 10m + 4) + 1


Here, remainder is 1 = .


Thus in any case, if a2 is divided by 6, then the remainder is 0, 1, 3 or 4 but not 5.

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