Q. 43 X5.0( 1 Vote )

# On division by 6, a2 cannot leave remainder (a N)A. 1B. 4C. 5D. 3

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

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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