a-n-a-n-1-2-a-n-2-2-If-a-1-1-and-a-2-1-what-is-the-remainder-of-a-2016-when-divided-by-10-

Question Number 9485 by Joel575 last updated on 10/Dec/16

a_{n} = a_{n - 1}^{2} + a_{n - 2}^{2}

If a_{1} = 1 and a_{2} = 1, what is the remainder of a_{2016} when divided by 10 ?

Commented by sou1618 last updated on 10/Dec/16

(m o d 10)

a_{3} = 1^{2} + 1^{2} = 2

a_{4} = 1^{2} + 2^{2} = 5

a_{5} = 2^{2} + 5^{2} = 29 \equiv 9

a_{6} = 5^{2} + 29^{2} \equiv 5^{2} + 9^{2} \equiv 5 + 1 = 6

a_{7} \equiv 9^{2} + 6^{2} \equiv 1 + 6 = 7

a_{8} \equiv 6^{2} + 7^{2} \equiv 6 + 9 \equiv 5

a_{9} \equiv 7^{2} + 5^{2} \equiv 9 + 5 \equiv 4

a_{10} \equiv 5^{2} + 4^{2} \equiv 5 + 6 \equiv 1

a_{11} \equiv 4^{2} + 1^{2} \equiv 6 + 1 = 7

a_{12} \equiv 1^{2} + 7^{2} \equiv 1 + 9 \equiv 0

a_{13} \equiv 7^{2} + 0^{2} \equiv 9 + 0 = 9

a_{14} \equiv 0^{2} + 9^{2} \equiv 0 + 1 = 1

a_{15} \equiv 9^{2} + 1^{2} \equiv 1 + 1 = 2

a_{2} \equiv a_{14}, a_{3} \equiv a_{15}

a_{16} = a_{14}^{2} + a_{15}^{2} \equiv a_{2}^{2} + a_{3}^{2} \equiv a_{4}

a_{17} = a_{15}^{2} + a_{16}^{2} \equiv a_{3}^{2} + a_{4}^{2} \equiv a_{5}

\dots .

(k = 1, 2, 3 \dots 10, 11, 12 / n = 0, 1, 2, 3 \dots . .)

a_{k + 12 n} \equiv a_{k}

a_{2016} = a_{12 + 12 \times 167} \equiv a_{12} = 0 (∵ 2016 = 12 \times 168)

a_{2016} \equiv 0 (m o d 10)

Commented by Joel575 last updated on 11/Dec/16

t h a n k y o u v e r y m u c h

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