17^(202) the end 3 number?


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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

	$17^{0} = 1$ $17^{1} = 17 \to 7$ $17^{2} : 7 \times 7 = 49 \to 9$ $17^{3} : 7 \times 9 = 63 \to 3$ $17^{4} : 7 \times 3 = 21 \to 1$ $. . .$ $The end digit is :$ $1 for 17^{4 n}, n \in N$ $7 for 17^{4 n + 1}$ $9 for 17^{4 n + 2}$ $3 for 17^{4 n + 3}$ $202 = 4 \times 50 + 2 \Rightarrow end digit is 9$ $Let u_{n} = \frac{17^{4 n + 2} - 9}{10}$ $u_{n + 1} = \frac{17^{4 (n + 1) + 2} - 9}{10}$ $u_{n + 1} = \frac{17^{4 .} 17^{4 n + 2} - 9}{10}$ $u_{n + 1} = \frac{17^{4 .} 10 . \frac{17^{4 n + 2} - 9}{10} + 9 . 17^{4} - 9}{10}$ $\frac{17^{4 .} 10 . u_{n} + 9 (17^{4} - 1)}{10}$ $End digit of 17^{4} = 1$ $\Rightarrow end digit of u_{n + 1} = end digit of u_{n}$ $u_{2} = 28 \Rightarrow end digit of u_{50} = 8$ $But u_{50} = \frac{17^{202} - 9}{10}$ $\Rightarrow second end digit of 17^{202} = 8$ $. . . to be continued . . .$
	Commented by MathSh last updated on 21/Dec/20

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