Q. 43 X5.0( 1 Vote )

# On division by 6,

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 a^{2} = 6^{2}m^{2}

= 36m^{2} = 6 × (6m)^{2} + 0

Here, remainder is 0.

If a = 6m + 1

Then a^{2} = (6m + 1)^{2}

= 36m^{2} + 12m + 1

= 6(6m^{2} + 2m) + 1

Here, remainder is 1.

If a = 6m + 2

Then a^{2} = (6m + 2)^{2}

= 36m^{2} + 24m + 4

= 6(6m^{2} + 4m) + 4

Here, remainder is 4.

If a = 6m + 3

Then a^{2} = (6m + 3)^{2}

= 36m^{2} + 32m + 9

= 6(6m^{2} + 6m + 1) + 3

Here, remainder is 3.

If a = 6m + 4

Then a^{2} = (6m + 4)^{2}

= 36m^{2} + 48m + 16

= 6(6m^{2} + 8m + 2) + 4

Here, remainder is 4.

If a = 6m + 5

Then a^{2} = (6m + 5)^{2}

= 36m^{2} + 60m + 25

= 6(6m^{2} + 10m + 4) + 1

Here, remainder is 1 = .

Thus in any case, if a^{2} is divided by 6, then the remainder is 0, 1, 3 or 4 but not 5.

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