Question

tHE NUMBER OF 5 DIGIT NUMBER DIVISIBLE BY 3 THAT CAN BE FORMED BY USING THE DIGITS 0 , 1 , 2,3,4,5 NOT REPEATING

Answer 1

@saifoo.khan @apoorvk @experimentX @Mani_Jha @Mertsj

Answer 2

@open_study1 what's the divisibility test for 3?

Answer 3

@open_study1 are you here?

Answer 4

YES NO IDEA

Answer 5

sum shd be divisible bye.....

Answer 6

3

Answer 7

WE ARE NOT PROVIDED WITH NUMBERS SO!!

Answer 8

u r provided :|

Answer 9

For testing a number to be divisible by 3 , the sum of digits should be divisible by 3
here we have 0,1,2,3,4, 5
let's sum= 1+2+4+5+0=12 ( this is divisible by 3)
We have two cases when 3 is a part of no. and when it's not
so if we have 3 included=1,2,3,4, 5
if we don't have 3=0,1,2,4,5
Now find the number of numbers for both the cases

Answer 10

I believe you use a permutation because the sum adds up to 18 which is divisible by 3, so you just needs to find out how many ways to arrange the numbers.

Answer 11

YES U R CORRECT

Answer 12

GUYS ANY IDEA

Answer 13

CAN ANYONE SOLVE THIS PLZZ

Answer 14

@ash2326 was trying to help you....@open_study1

Answer 15

@open_study1 did you get my point?

Answer 16

YES

Answer 17

We found that sum of 1,2,4,5 is 12
so either 3 or 0 should be included , cause if we don't include any other numeral, the sum would not be divisible by 3
so we have two cases
1) When 3 is included
so we have digits 1,2,3,4,5
Now we need to find the number of numbers (without repeating)
=
\[\huge -----\]
first place can be filled in 5 ways, second can be filled in 4 and so on
\[5\times 4\times 3\times 2\times 1=120\]
Do you get this part?

Answer 18

SO WHO IS CORRECT @eliassaab OR @ash2326 ???? CONFUSED

Answer 19

@eliassaab it's not mentioned that the first digit should be 5, the only condition is that the number should be divisible by 3

Answer 20

THE NUMBER SHOULD BE DIVISIBLE BY3 AND THE DIGITS SHOULD NOT REPEAT

Answer 21

Yeah so did you get my last post?

Answer 22

Sorry, I thought divisible by 5.

Answer 23

LET ME CHECK ONCE AGAIN

Answer 24

@open_study1 why are you using caps?

Answer 25

@ZhangYan is that disturbs u i am sorry

Answer 26

no no its fine...i'm just asking

Answer 27

ok

Answer 28

@eliassaab
if 3 is not included and we use 0 instead
then we have 0, 1, 2, 4, 5
so we would have
\[4 \times 4 \times 3\times 2\times 1=96\]
96 more
total numbers
\[120+96=216\]

Answer 29

@ash2326 u r correct thanzzz

Answer 30

Here are the 216 numbers
{10245, 10254, 10425, 10452, 10524, 10542, 12045, 12054, 12345, \
12354, 12405, 12435, 12450, 12453, 12504, 12534, 12540, 12543, 13245, \
13254, 13425, 13452, 13524, 13542, 14025, 14052, 14205, 14235, 14250, \
14253, 14325, 14352, 14502, 14520, 14523, 14532, 15024, 15042, 15204, \
15234, 15240, 15243, 15324, 15342, 15402, 15420, 15423, 15432, 20145, \
20154, 20415, 20451, 20514, 20541, 21045, 21054, 21345, 21354, 21405, \
21435, 21450, 21453, 21504, 21534, 21540, 21543, 23145, 23154, 23415, \
23451, 23514, 23541, 24015, 24051, 24105, 24135, 24150, 24153, 24315, \
24351, 24501, 24510, 24513, 24531, 25014, 25041, 25104, 25134, 25140, \
25143, 25314, 25341, 25401, 25410, 25413, 25431, 31245, 31254, 31425, \
31452, 31524, 31542, 32145, 32154, 32415, 32451, 32514, 32541, 34125, \
34152, 34215, 34251, 34512, 34521, 35124, 35142, 35214, 35241, 35412, \
35421, 40125, 40152, 40215, 40251, 40512, 40521, 41025, 41052, 41205, \
41235, 41250, 41253, 41325, 41352, 41502, 41520, 41523, 41532, 42015, \
42051, 42105, 42135, 42150, 42153, 42315, 42351, 42501, 42510, 42513, \
42531, 43125, 43152, 43215, 43251, 43512, 43521, 45012, 45021, 45102, \
45120, 45123, 45132, 45201, 45210, 45213, 45231, 45312, 45321, 50124, \
50142, 50214, 50241, 50412, 50421, 51024, 51042, 51204, 51234, 51240, \
51243, 51324, 51342, 51402, 51420, 51423, 51432, 52014, 52041, 52104, \
52134, 52140, 52143, 52314, 52341, 52401, 52410, 52413, 52431, 53124, \
53142, 53214, 53241, 53412, 53421, 54012, 54021, 54102, 54120, 54123, \
54132, 54201, 54210, 54213, 54231, 54312, 54321}