17-202-the-end-3-number-

Question Number 126542 by MathSh last updated on 21/Dec/20

17^{202}

t h e e n d 3 n u m b e r ?

Commented by AlagaIbile last updated on 21/Dec/20

Answered by Olaf last updated on 21/Dec/20

17^{0} = 1

17^{1} = 17 \to 7

17^{2} : 7 \times 7 = 49 \to 9

17^{3} : 7 \times 9 = 63 \to 3

17^{4} : 7 \times 3 = 21 \to 1

\dots

The end digit is :

1 for 17^{4 n}, n \in N

7 for 17^{4 n + 1}

9 for 17^{4 n + 2}

3 for 17^{4 n + 3}

202 = 4 \times 50 + 2 \Rightarrow end digit is 9

Let u_{n} = \frac{17^{4 n + 2} - 9}{10}

u_{n + 1} = \frac{17^{4 (n + 1) + 2} - 9}{10}

u_{n + 1} = \frac{17^{4 .} 17^{4 n + 2} - 9}{10}

u_{n + 1} = \frac{17^{4 .} 10 . \frac{17^{4 n + 2} - 9}{10} + 9 . 17^{4} - 9}{10}

\frac{17^{4 .} 10 . u_{n} + 9 (17^{4} - 1)}{10}

End digit of 17^{4} = 1

\Rightarrow end digit of u_{n + 1} = end digit of u_{n}

u_{2} = 28 \Rightarrow end digit of u_{50} = 8

But u_{50} = \frac{17^{202} - 9}{10}

\Rightarrow second end digit of 17^{202} = 8

\dots to be continued \dots

Commented by MathSh last updated on 21/Dec/20

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