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Question Number 191758 by Mingma last updated on 30/Apr/23

	Answered by deleteduser1 last updated on 30/Apr/23

	$521^{1000} \equiv 0001 (m o d 10000)$ $(521, 10000) = 1 \Rightarrow 521^{999} \equiv \frac{1}{521} (m o d 10000)$ $\frac{1}{521} \equiv x (m o d 10000) \Rightarrow 521 x \equiv 1 (m o d 10000)$ $10000 q - 521 x = - 1$ $\Rightarrow x i s o d d$ $521 x = 521 (2 k + 1) = (500 + 21) (2 k + 1)$ $= 500 (2 k + 1) + 21 x = 1000 k + 500 + 21 x$ $\Rightarrow 521 x \equiv 500 + 21 x \equiv 1 (m o d 10000)$ $\Rightarrow 21 x \equiv - 499 \equiv 9501 (m o d 10000)$ $21 x \equiv 9501 (m o d 10000) \Rightarrow 21 x \equiv 39501 (m o d 10000)$ $\Rightarrow x \equiv 1881 (m o d 10000)$ $\Rightarrow L a s t f o u r d i g i t s o f 521^{999} = 1881$
	Commented by malwan last updated on 30/Apr/23

	$1881$
	Commented by Mingma last updated on 30/Apr/23
	How did you jump from 10000q-521x-=1 to 21x modulo 9501? Can you explain that, please!
	Commented by deleteduser1 last updated on 30/Apr/23

	$D o n e .$
	Commented by Mingma last updated on 30/Apr/23
	Perfect ��
	Answered by mr W last updated on 30/Apr/23

	$k n o w n : 521^{1000} = 10^{4} n + 1$ $s a y : 521^{999} = 10^{4} m + x w i t h 1 ⩽ x < 10^{4}$ $(10^{4} m + x) \times 521 = 10^{4} n + 1$ $\Rightarrow 521 x - 1 = (n - 521 m) \times 10^{4} = k \times 10^{4}$ $w e s e e t h a t l a s t d i g i t o f x m u s t b e 1 .$ $s a y x = 10 y + 1, t h e n$ $521 (10 y + 1) - 1 = k \times 10^{4}$ $\Rightarrow 521 y + 52 = k \times 10^{3}$ $w e s e e t h e l a s t d i g i t o f y m u s t b e 8,$ $i . e . t h e l a s t 2 d i g i t s o f x m u s t b e 81 .$ $s a y x = 100 z + 81, t h e n$ $521 (100 z + 81) - 1 = k \times 10^{4}$ $\Rightarrow 521 z + 422 = k \times 10^{2}$ $w e s e e t h e l a s t d i g i t o f z m u s t b e 8,$ $i . e . t h e l a s t 3 d i g i t s o f x m u s t b e 881 .$ $s a y x = 1000 w + 881, t h e n$ $521 (1000 w + 881) - 1 = k \times 10^{4}$ $\Rightarrow 521 w + 459 = k \times 10$ $w e s e e t h e l a s t d i g i t o f w m u s t b e 1,$ $i . e . t h e l a s t 4 d i g i t s o f x m u s t b e 1881 .$ $\Rightarrow x = 1881$ $i . e . t h e l a s t f o u r d i g i t s o f 521^{999} a r e$ $1881 .$
	Commented by Mingma last updated on 30/Apr/23
	Nice work!
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