How many five digit numbers can be formed using digits 0, 1, 2, 3, 4

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.

so total 120+96=216

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).

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.

Hope it's clear.

0,1,2,3,4,5

One digit will have to remain out for all 5 digit numbers;

if 0 is out; Leftover digits will be 1,2,3,4,5 = Sum(1,2,3,4,5)=15.
5! = 120 numbers

if 1 is out; Leftover digits will be 0,2,3,4,5 = Sum(0,2,3,4,5)=14. Ignore(Not divisible by 3)

if 3 is out; Leftover digits will be 0,1,2,4,5 = Sum(0,1,2,4,5)=12.
4*4! = 4*24 = 96

if 4 is out; Leftover digits will be 0,1,2,3,5 = Sum(0,1,2,3,5)=11. Ignore
if 5 is out; Leftover digits will be 0,1,2,3,4 = Sum(0,1,2,3,4)=10. Ignore

Total count of numbers divisible by 3 = 120+96 = 216

Ans: "E"

A number is divisible by 3 if sum of its digits is a multiple of 3.

With the given set of digits, there are two possible combinations of 5 digits each-

A. [1,2,3,4,5] No. of possible 5 digit numbers: 5!= 120
B. [0,1,2,4,5] No. of possible 5 digit numbers: 4*4!=96 [the number can't start with a 0]

A+B= 120+96= 216

E

0 + 1 + 2 + 3 + 4 + 5 = 15

To form 5-digit number, we can remove a digit and the sum should still be divisible by 3.

15 - 1 = 14
15 - 2 = 13
15 - 3 = 12 BINGO!
15 - 4 = 11
15 - 5 = 10

Possible = {5,4,3,2,1} and {5,4,0,2,1}

There are 5! = 120 ways to arrange {5,4,3,2,1}
There are 5! - 5!/5 = 96 ways to arrange {5,4,0,2,1} since 0 cannot start the five number digit.

120 + 96 = 216

Answer: E

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...

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.

I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.

What is wrong with this logic?

Because the numbers divisible by 3 are not 1/3rd of all possible numbers.

{0, 1, 2, 3, 4} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.

Total = 5*5*4*3*2 = 600 but the numbers which are divisible by 3 come from third and sixth sets: 96 + 120 = 216.

We cannot do this because we have the asymmetric 0 as one of the digits. The number of 5 digit numbers that can be formed with 0, 1, 2, 3 and 4 is different from the number of 5 digit numbers that can be formed with 1, 2, 3, 4 and 5 (because 0 cannot be the first digit).

Had the digits been 1, 2, 3, 4, 5 and 6, then your method would have been correct.

If 0 is included:
{0, 1, 2, 3, 4} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set. - All these numbers are divisible by 3
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
The number of 5 digit numbers in these sets is not the same - Sets with 0 have fewer numbers

If 0 is not included:
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 4, 6} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
{1, 2, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{1, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{2, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
Here exactly 1/3rd of the numbers will be divisible by 3.

We need to form five digit numbers with distinct digits which are divisible by 3. A number is divisible by 3 when the sum of its digits is divisible by 3.

When we observe the digits, we see that the sum of the given digits is 15. Typical of a GMAT kind of question – data given is very precise and rarely vague.

So, there are only two cases that can be considered to fit the constraints given

Case 1: A 5 digit number with the digits {1,2,3,4,5}. Since 0 is not a part of this set and there are 5 different digits, we can form a total of 5P5 = 5! = 120 numbers. All of these will be divisible by 3.

At this stage, we can eliminate answer options A, B and C.

Case 2: A 5 digit number with the digits {0,1, 2, 4, 5}. Since 0 is a part of this set, we need to use Counting methods to find out the number of 5-digit numbers.

The ten thousands place can be filled in 4 ways, since 0 cannot come here; the thousands place can be filled in 4 ways, the hundreds in 3 ways, the tens in 2 ways and the units place in 1 way.
Therefore, number of 5-digit numbers = 4 * 4 * 3 * 2 * 1 = 96.

Total number of 5-digit numbers with distinct digits, divisible by 3 = Case 1 + Case 2 = 120 + 96. Answer option D can be eliminated.

The correct answer option is E.

Hope that helps!
Aravind B T

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