banerjeea_98 wrote:

What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6

Let A = 9^9 + 9^8 + ... + 9^1

9A = 9^10 + (9^9 + 9^8 + ... + 9^2 + 9^1) - 9^1

9A = 9^10 + A - 9

8A = 9^10 - 9

A = (9^10 - 9)/8

= 9( 9^9 -1)/8

Since 9^9 = (9^3)^3

A = 9 ((9^3)^3 - 1)/8

= 9 (729^3 -1)/8

Since a^3 - b^3 = (a-b)^3 + 3ab(a-b)

A = 9/8[(729-1)^3 + 3*729(729-1)]

= 9/8[728^3 + 3*729*728]

= 9[91*728^2 + 3*729*91]

Dividing this by 6, we have this remainder

9[ 1*2*2 + 3*3*1 ] = 9*13 = 117 which leaves 3 as the remainder.

Thus I'd go with 3 as the answer.

Hope that helps.

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