Find the number of all such 7−digit no. which satisfy conditions: 1) divisible by 3 2) repetition allowed 3) zero is not used.


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Question Number 49408 by rahul 19 last updated on 06/Dec/18

$F i n d t h e n u m b e r o f a l l s u c h 7 - d i g i t$ $n o . w h i c h s a t i s f y c o n d i t i o n s :$ $1) d i v i s i b l e b y 3$ $2) r e p e t i t i o n a l l o w e d$ $3) z e r o i s n o t u s e d .$
Commented by mr W last updated on 06/Dec/18

$t h e r e a r e o n l y t h r e e k i n d s o f n u m b e r s :$ $3 n + k w i t h k = 0, 1, 2$ $w e n e e d o n l y t h e n u m b e r s w i t h k = 0 .$ $i . e . o n l y a t h i r d o f 9^{7} n u m b e r s c a n b e$ $d i v i d e d b y 3, s o t h e a n s w e r i s$ $\frac{9^{7}}{3} = \frac{4782969}{3} = 1594323$
Commented by Kunal12588 last updated on 06/Dec/18

$t h e r e a r e 9^{7} = 4782969 w h i c h s a t i s f i e s 2^{n d}$ $a n d 3^{r d} c o n d i t i o n t h e p r o b l e m i s 1^{s t} c o n d i t i o n$
Commented by rahul 19 last updated on 06/Dec/18
Yes, the problem is tough. As all three conditions must be satisfied simultaneously!
	Answered by MJS last updated on 06/Dec/18

	$all 6 - digit numbers without zero$ $9^{6} = 531441$ $each \frac{1}{3} of them divisible by 3 with remainers 0, 1, 2$ $add 3, 2, 1 as 7^{th} digit and {they}^{'} re all divisible$ $by 3 with remainder 0$ $\Rightarrow 531441 numbers satisfy the conditions$
	Commented by MJS last updated on 06/Dec/18

	$1111113 1111212 . . .$ $1111122 1111221$ $1111131 1111233$ $1111143 1111242$ $1111152 1111251$ $1111161 1111263$ $1111173 1111272$ $1111182 1111281$ $1111191 1111293$
	Commented by mr W last updated on 06/Dec/18

	$v e r y n i c e s i r! b u t t h e a n s w e r s h o u l d b e$ $531441 \times 3 = 1594323$ $1111113 / 6 / 9$ $1111122 / 5 / 8$ $1111131 / 4 / 7$ $1111143 / 6 / 9$ $. . . . . .$
	Commented by MJS last updated on 06/Dec/18

	${you}^{'} re absolutely right! I had been in a hurry$ $but anyway thought the number of these$ $numbers is too small . . .$
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