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Re: digit counting [#permalink]
22 Oct 2009, 13:59

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TheRob wrote:

How many five digit numbers can be formed using digits 0,1,2,3,4,5, Which are divisible by 3, whithout any of the digits repeating?

A. 15 B. 96 C. 120 D. 181 E. 216

First step: We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0,1,2,3,4,5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers: {1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:

{1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

{0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96

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?

(C) 2008 GMAT Club - m04#32

15 96 120 181 216

HOW DO YOU SOLVE THIS, I CAN'T UNDERSTAND IT

For a number to be divisible by 3 sum of its digits should be divisible by 3, so in 0,1,2,3,4,5 set of digits that can be together 1,2,3,4,5, and 0,1,2,4,5 (drop one digit at a time and sum rest of the others to find this ) in first set numbers that can be there = 5*4*3*2*1 in second set = 4*4*3*2*1 (since zero cannot be the last digit)

total = 5*4*3*2*1 + 4*4*3*2*1 = 4! * (5+4) = 24 * 9 = 216, hence 216 is the answer. _________________

By the property of divisibility by 3 i.e "a no: is divisible by 3, if the sum of the digits is divisible by 3"(e.g= 12-->1+2=3)

so from 0,1,2,3,4,5 the set of 5 digit no:s that can be formed which is divisible by 3 are 0,1,2,4,5(sum=12) & 1,2,3,4,5(sum=15)

from first set(0,1,2,4,5) no:s formed are 96 i.e first digit can be formed from any 4 no: except 0, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.

from second set(1,2,3,4,5) no:s formed are 120 i.e first digit can be formed from any 5 digits, second digit from 4 no: except digit used at first place,3rd from rest 3 , 4th from rest 2 no: and in fifth remaining digit since no repetition allowed.

Re: digit counting [#permalink]
06 Feb 2011, 13:48

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks _________________

Re: digit counting [#permalink]
06 Feb 2011, 13:55

Expert's post

ajit257 wrote:

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

i understood the first part but did not get the second part 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3. ..Could you please explain it in a little bit more detail. Thanks

The sum of the given digits is already a multiple of 3 (15), in order the sum of 5 digits to be a multiple of 3 you must withdraw a digit which is itself a multiple of 3, otherwise (multiple of 3) - (non-multiple of 3) = (non-multiple of 3). _________________

Re: digit counting [#permalink]
06 Feb 2011, 14:18

Expert's post

ajit257 wrote:

so lets say we were asked a multiple of 5 so in that case we would have to withdraw the digit 5 ..is that correct ?

5 or 0, as 0 is also a multiple of 5.

AGAIN: we have (sum of 6 digits)=(multiple of 3). Question what digit should we withdraw so that the sum of the remaining 5 digits remain a multiple of 3? Answer: the digit which is itself a multiple of 3.

Below might help to understand this concept better.

If integers a and b are both multiples of some integer k>1 (divisible by k), then their sum and difference will also be a multiple of k (divisible by k): Example: a=6 and b=9, both divisible by 3 ---> a+b=15 and a-b=-3, again both divisible by 3.

If out of integers a and b one is a multiple of some integer k>1 and another is not, then their sum and difference will NOT be a multiple of k (divisible by k): Example: a=6, divisible by 3 and b=5, not divisible by 3 ---> a+b=11 and a-b=1, neither is divisible by 3.

If integers a and b both are NOT multiples of some integer k>1 (divisible by k), then their sum and difference may or may not be a multiple of k (divisible by k): Example: a=5 and b=4, neither is divisible by 3 ---> a+b=9, is divisible by 3 and a-b=1, is not divisible by 3; OR: a=6 and b=3, neither is divisible by 5 ---> a+b=9 and a-b=3, neither is divisible by 5; OR: a=2 and b=2, neither is divisible by 4 ---> a+b=4 and a-b=0, both are divisible by 4.

Re: How many five-digit numbers can be formed using digits [#permalink]
13 Jan 2013, 23:40

I did in 1 min 18 sec. At first I wanted to choose a set of five digits, but started to worry about the complications with the leading zero.

Then I thought that the last digits could always be chosen in only two ways so as to ensure divisibility by three - however, I quickly realized that I would not get all different digits.

Then I realized that once I get a number I can keep permuting the digits while still getting valid numbers.

In an attempt to avoid the leading zero I tried 12345 and noticed that it was divisible by 3. Thus, I've got 5!=120 answers and immediately eliminated two answers, A and B.

Then I addressed the case of a leading zero. Since I wanted to preserve divisibility by 3, I quickly saw that I could only use 0 instead of 3. Thus, the only other possible set was {0, 1, 2, 4, 5}. I tried adding another 5! and got 240, so the answer was slightly less than that.

After that I knew I had to subtract 4!=24 to account for all the possibilities with a leading zero, which left me with 240-24=216. This is how I do such problems... _________________

Re: How many five-digit numbers can be formed using digits [#permalink]
15 Oct 2013, 17:26

E.

I'm offering up a way to apply the "slot method" to this problem below.

First, as everyone else has identified above, you need to find the cases where the 6 numbers (0, 1, 2, 3, 4, 5) create a 5-digit number divisible by 3.

Shortcut review: a number is divisible by 3 if the sum of the digits in the number is divisible by 3.

So, 12345 would be divisible by 3 (1+2+3+4+5 = 15, which is divisible by 3).

Analyzing the given numbers, we can conclude that only the following two groups of numbers work: 1, 2, 3, 4, 5 (in any order, they would create a five digit number divisible by 3 - confirmed by the shortcut above), and 0, 1,2, 4, 5.

Now we need to count the possible arrangements in both cases, and then add them together.

To use the slot method with case 1 (1,2,3,4,5): _ _ _ _ _ (five digit number, 5 slots). Fill in the slots with the number of "choices" left over from your pool of numbers. Starting from the left, I have 5 choices I can put in slot #1 (5 numbers from the group 1,2,3,4,5 - pretend I put in number 1, that leaves 4 numbers) 5 _ _ _ _ Fill in the next slot with the number of choices left over (4 choices left...numbers 2 through 5) 5 4 _ _ _ Continue filling out the slots until you arrive at: 5 4 3 2 1 Multiply the choices together: 5x4x3x2x1 (which also happens to be 5!) = 120 different arrangements for the first case.

Now consider case 2 (0,1,2,4,5): _ _ _ _ _ (five digit number, 5 slots). Here's the tricky part. I can't put 0 as the first digit in the number...that would make it a 4-digit number! So I only have 4 choices to pick from for my first slot! 4 _ _ _ _ Fill in the next slot with the remaining choices (if I put in 1 in the first slot, I have 0,2,4,5 left over...so 4 more choices to go). 4 4 _ _ _ Continue to fill out the slots with the remaining choices: 4 4 3 2 1 Multiply the choices together: 4 x 4! = 96 different arrangements for the second case.

Now the final step is to add all the possible arrangements together from case 1 and case 2: 120 + 96 = 216. And this is our answer.

Hope this alternate "slot" method helps! This is how I try to work these combinatoric problems instead of using formulas... in this case it worked out nicely. Here, order didn't matter (we are only looking for total possible arrangements) in the digits, so we didn't need to divide by the factorial number of slots.

gmatclubot

Re: How many five-digit numbers can be formed using digits
[#permalink]
15 Oct 2013, 17:26