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!
