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by rijul007 » Sat Mar 31, 2012 5:45 am
Md.Nazrul Islam wrote:How many 5 digit numbers can be formed which are divisible by 3using the numbers 0,1,2,3,4,5 with repetition and repetition .
divisibility rule of 3 = sum of digits divisible by 3

1+2+3+4+5 =15 [divisible by 3]
no of 5 digit nos = 5! = 120

0+1+2+4+5 = 12 [divisible by 3]
no of 5 digit nos = 4*4*3*2*1 = 96

Total = 120 + 96 = 216
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by Pharo » Sat Mar 31, 2012 6:29 am
rijul007, even though I can not get the right answer; your answer is wrong dude. For a start, you do not consider repetition in your answer. (i.e : 55500 is not considered in your answer.) Also what about 10002 or 11100? They are divisible by 3 but they are not included in your answer or 50001; digits sum up to 6 and thus divisible by 3 and so on..

It is a good start though and I like your signature!

Md.Nazrul Islam : I have been working on this for a while and I cannot figure out a way to quickly solve this. When are you planning on posting the solution? (And please do not say the only way to solve it is by doing a case by case analysis :P )
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by rijul007 » Sat Mar 31, 2012 9:42 am
Pharo wrote:rijul007, even though I can not get the right answer; your answer is wrong dude. For a start, you do not consider repetition in your answer. (i.e : 55500 is not considered in your answer.) Also what about 10002 or 11100? They are divisible by 3 but they are not included in your answer or 50001; digits sum up to 6 and thus divisible by 3 and so on..

It is a good start though and I like your signature!
oh yes, sorry :D i missed the repititions
my bad


this can be solved by making cases.. it would be lengthy and frustrating and i dont feel like doing this..
i don't think there can be any other quicker way to slve this... (i might even be wrong :D )

if there really isn't .. then there is no way we would find this ques on the GMAT...
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by Pharo » Sat Mar 31, 2012 9:58 pm
This feels like the kind of question a math teacher would ask in high school to screw the students :P

I will keep checking here for an answer :)
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by Bill@VeritasPrep » Sat Mar 31, 2012 10:22 pm
Repetition makes this one brutal. By looking for digits that add to a multiple of 3, here's what I have so far:

Sum=24
55554

Sum=21
55551
55533
55542
55443
44445

Sum=18
55530
55521
55440
54432
44442
44433
44451

Sum=15
55500
55320
55410
44430
44421
54420
54411

Now, for each of those, you would run the permutations with repeating elements formula. Oh, and the ones that involve a 0? You have to subtract all of the permutations that start with 0 (or two 0's in the case of 55500).
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by Pharo » Sat Mar 31, 2012 10:37 pm
Hehe yeah I have done a similar sort of start as well :)

Seems like collectively we can write down all the cases :P
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