can you find any arrangement of nine digits of 1−9 such as 967854312 and the first digit should be divisible by 1 thefirst two digitds should be divisible by 2 the first three digitds should be divisible by 3 ...... the number should be fivisible by 9 if is such a number available


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Question Number 208555 by liuxinnan last updated on 18/Jun/24

$c a n y o u f i n d a n y a r r a n g e m e n t o f n i n e d i g i t s o f 1 - 9$ $s u c h a s 967854312$ $a n d t h e f i r s t d i g i t s h o u l d b e d i v i s i b l e b y 1$ $t h e f i r s t t w o d i g i t d s s h o u l d b e d i v i s i b l e b y 2$ $t h e f i r s t t h r e e d i g i t d s s h o u l d b e d i v i s i b l e b y 3$ $. . . . . .$ $t h e n u m b e r s h o u l d b e f i v i s i b l e b y 9$ $i f i s s u c h a n u m b e r a v a i l a b l e$
	Answered by nikif99 last updated on 18/Jun/24

	$5 t h d i g i t i s 5 .$ $2 n d, 4 t h, 6 t h, 8 t h d i g i t s a r e e v e n .$ $1 s t, 3 r d, 7 t h, 9 t h d i g i t s a r e o d d .$ $4 t h i s 2 o r 6 (e n d i n g i n 12, 16, 32, 36 . . .)$ $(1 s t + 2 n d + 3 r d) a s w e l l (4 t h + 5 t h + 6 t h)$ $m u s t b e d i v i d e d b y 3 .$ $. . .$ $B y e l i m i n a t i n g s o m e c a s e s a n d ‘ ‘ t r y$ $a n d {c h e c k}^{″}, y o u r e a c h 381654729 .$ $P S . T h e n, y o u a d d a 0 a n d y o u o b t a i n$ $a n u m b e r d u v i s i b l e b y 10 .$
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