rupsk wrote:Pls help me in solving below problem

How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)

A. 216
B. 3152
C. 240
D. 600
E. 305

thanks

Hi!

First, we need to know a trick for mutliples of 3: the sum of the digits of any multiple of 3 will also be a multiple of 3.

So, for example, 54321 is a multiple of 3, since 5+4+3+2+1=15 and 15 is a multiple of 3.

Now we need to see which combinations of those 5 digits will sum to a multiple of 3.

12345 works (as shown above).
12045 works (1+2+4+5 = 12).

That's it!

We know that we're done because in order to generate another multiple of 3, we'd have to replace one of our digits with another digits that's exactly 3 higher or lower than the original digit. Since we started with 12345, the only "legal" swap we can make is 3 for 0.

We also need to be careful when counting the possible permutations of our sets.

For 12345, there are simply 5! = 120 possible arrangements.

however, for 01234, we have to remember that we're creating a 5 digit number, so 0 can't hold the first place (since 0 in the first spot wouldn't be a significant digit).

So, our first slot can be 4 different choices, then 4 choices for slot 2, 3 for slot 3, 2 for slot 4 and 1 for slot 5, giving us:

4*4*3*2*1 = 96 possibilities.

We want the total number of possibilities, so we add:

120 + 96 = 216: choose (A).

* * *

As an aside, please post problem solving questions in the problem solving sub-forum!

rupsk wrote:Pls help me in solving below problem

How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0, 1, 2, 3, 4, 5 (WITHOUT REPETITION)

A. 216
B. 3152
C. 240
D. 600
E. 305

thanks

a number is divisible by 3 when sum of its digit is divisible by 3.

now here two cases are possible..!!

case 1) when 0 is selected as one of the number while forming a five digit numbers which are divisible by 3.
case 2) when 0 is not selected.

case 1) - - - - -; here five possible number that will form a 5 digit number divisible by 3 would be
01245; the left most digit can be filled by any of 1,2,4,5 in 4 ways, (we can't have 0 in the left most position because that will result in making the five digit number as a four digit number.);
and the remaining 4 digits can arrange themselves at the remaining 4 blank spaces in 4! ways.

hence required no. of ways= 4*4!=96;

case 2) when 0 is not included in the group then we have 5 possible no. as 12345; these 5 numbers can arrange themselves in 5!;
5!=120;

total no. of ways = 96+120=216 hence A

Thanks. I also got same answer but they had given 241 as correct answer. So i got confused but thanks you for response.