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Hello does anyone has a clue on this one I thought 40 was the answer but i was wrong plz Explains your works and reasoning thanks

How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? (1) 15 2(40) (3) 96 (4) 216 (5) 120

Topic locked. In case of any questions please continue the discussion here: ps-91597.html

Last edited by Bunuel on 13 Jan 2012, 06:19, edited 2 times in total.

Topic locked. In case of any questions please continue the discussion here: http://gmatclub.com/forum/ps-91597.html

I followed qpoo's logic
Only the following sets of numbers add up to a number that is divisible by 3
{54321} and (54210}

The answer is the factorial of the first set plus a little less than the factorial of the second set. If 0 is in the first digit from the left then you don't really have a 5 digit number do you.

This first set is 5! = 120. At this point you can solve the problem because there is only one answer larger than 120.

If you want to take the problem farther than when 0 is all the way to the left you have 4! so my answer is 2*5! - 4! = 216

0 can't be the leftmost digit, otherwise it will be a four digit number
So total numbers = 5! = 120
Numbers with 0 as leftmost digit = 4! =24
So total 5 digit numbers = 120 - 24 =96
Total 5 digit numbers = Total numbers (non 0) + Total numbers (including 0 but not at first place) = 120 + 96 =216

we can have two scenarios with a sum that is divisible by 13.

sum of 15
12345

or

sum of 12
01245

We can permute the first scenario in 5! ways.
5! = 120

We can permute the second scenario in 4*4! ways.
The first slot cannot be zero. Remove zero from the options. 4 ways to fill first slot. Replace zero in the second slot and account for the number we placed in the first slot. We have 5-1+1= 4 or 4! to permute the rest of the slots.

Hello does anyone has a clue on this one I thought 40 was the answer but i was wrong plz Explains your works and reasoning thanks

How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? (1) 15 2(40) (3) 96 (4) 216 (5) 120

= non zero -5 digit + include zerio in five digit ={12345} +{01245}

= 5! + 4*4*3*2 = 120+96 =216
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I'd think its E - 120 Explanation: # of 5 digit numbers using (0-5) = 6P5 # of the above, that has 0 in the beginning = 6P4

Total legit 5-digit numbers = 6P5- 6P4 = 360

I kind of at this point, reckoned a third of these must be divisible by 3 = 120 (Can someone give a better way of adding in divisibility by 3 as a criteria?)

RULE: When sum of digits of a number is divisible by 3, that number is divisible by 3. In 0,1,2,3,4,5, we can form 5 digits out of 1,2,3,4,5 and 0,1,2,4,5 that are divisible by 3.

1. 1,2,3,4,5 -> 5! = 120 2. 0,1,2,4,5 -> 5! = 120. However, can't have leftmost digit = 0. So, numbers to be removed = 4! = 24 -> 120 - 24 = 96

Total = 120 + 96 = 216 -> D.
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