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

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 \dots

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

\dots \dots

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 \dots

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