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

Question

How many positive five-digit multiples of 3 can be formed using the digits 0, 1, 2, 3, 4, and 5, without repeating any digit?

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

Bunuel · Accepted Answer

Official Solution:

How many positive five-digit multiples of 3 can be formed using the digits 0, 1, 2, 3, 4, and 5, without repeating any digit?

A. 96
B. 120
C. 181
D. 200
E. 216

First step:

Our goal is to determine which 5 digits out of the given 6 will form a 5-digit number divisible by 3.

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

For a number to be divisible by 3, the sum of its digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3), the only 5-digit combinations that would yield a number divisible by 3 are 15-0 = {1, 2, 3, 4, 5} and 15-3 = {0, 1, 2, 4, 5}. No other combinations of 5 digits from the given 6 will result in a number divisible by 3.
 
 Second step:

We have two sets of numbers:

{1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. Let's find out how many 5-digit numbers can be formed using these two sets:

{1, 2, 3, 4, 5} This set provides 5! = 120 numbers, as any permutation of these digits would result in a 5-digit number divisible by 3.

{0, 1, 2, 4, 5}. In this case, we cannot use 0 as the first digit, otherwise the number will no longer be a 5-digit number but a 4-digit number. Therefore, the desired number of combinations is the total combinations 5! minus the combinations with 0 as the first digit (permutations of the remaining 4 digits) 4!: 5! - 4! = 120 - 24 = 96.

In total, there are 120 + 96 = 216 possible 5-digit numbers that are divisible by 3.

Answer: E

deepakraam · Answer

Only 2 sets are possible

case (1) 1,2,3,4,5
case (2) 0,1,2,4,5.

case (1) : there will 5! ways to form the number = 120

case (2) ; there will 4*4*3*2*1 = 96 ways

So total no.of ways = 120+96 = 216 ways

LM · 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.

so total 120+96=216

ajit257 · Answer

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

Bunuel · Answer

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).

ajit257 · Answer

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 ?

Bunuel · Answer

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.

fluke · Answer

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"

blink005 · Answer

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.   No. of possible 5 digit numbers: 5!= 120
B.   No. of possible 5 digit numbers: 4*4!=96

A+B= 120+96= 216

E

mbaiseasy · Answer

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

SergeyOrshanskiy · Answer

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...

rltw · Answer

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.

ronr34 · Answer

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?

Bunuel · Answer

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.

KarishmaB · Answer

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.

CrackverbalGMAT · Answer

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

harryzhang · Answer

This is a very good question - it tests a few things that are easy to miss:

1) for a 5-digit abcde to be divisible by 3, a + b + c + d + e has to be divisible by 3

we can see that since only 1+2, 4+5, 0, and 3 would work, we can conclude that only two pairings are possible:
- 1, 2, 4, 5, 0
- 1, 2, 4, 5, 3

2) it's very tempting to just do 5! + 5! = 240 (my gut reaction as well), and im sure some questions will have 240 as a choice just to mess with us, but it's always important to check for exceptions (*note: a common one is sth. like a 5-digit number, and 0 is usually a good signal for a trap)

for a number to be five-digit, we cannot have 01245 or 02154 (i.e., 0 cannot be in the ten-thousands digit)

- 1, 2, 4, 5, 3 -> 5! = 120
- 1, 2, 4, 5, 0 -> how?

two ways: 
(1) backward
ALL - exception = all cases - where 0 is at the ten-thousands digit = 5! - 4! = 96
why 4!? you can think of the exception case as 0 being fixed and the rest (1, 2, 4, 5) can be arranged freely, and order matters, so permutation

(2) forward
first digit has 4 options (1, 2, 4, 5) and second digit has 4 options again (0 and 3 remaining ones), .... so 4 x 4 x 3 x 2 x 1 = 4 x 4! = 96

3) we then need to consider, do we multiply or add the two possibilities 
a good way to determine (or double-check this) is to assess whether they can both exist in the same answer: if yes, then multiply, and if no, then add them together
here, we cannot have both 0 and 3 in the same answer, so we should add them together

120 + 96 = (E) 216

BONUS QUESTION: how many six-digit numbers can be formed using digits 0,0,2,2,3,3?
  
STEPS:
(1) begin with writing out a few examples
(2) identify the restrictions / special requirements (e.g., six-digit numbers)
(3) apply formula; if there's a special requirement, tackle it FIRST

so we cannot have 002233 or 003232... AND 220033 and 220033 are the same (2 and 3 have duplicates) -> order doesnt matter for those

here, we would do similar things:

identify that the first digit can only be 2 or 3, so we have:

2 _ _ _ _ AND 3 _ _ _ _

for 2...., we have 0 0 3 3 2 to be freely arranged -> 5!/2!2!
for 3...., we have 0 0 3 2 2 to be freely arranged -> 5!/2!2!

so we add them together: 5!/2!2! + 5!/2!2! = 60

WiziusCareers1 · Answer

To solve this, we must apply the divisibility rule for 3: a number is a multiple of 3 if and only if the sum of its digits is divisible by 3.

Explanation 1: Short and Quick 
Select Digit Sets: Out of the six available digits {0,1,2,3,4,5}, the sum is 15. To form a five-digit number without repetition, we must exclude one digit. The excluded digit must be a multiple of 3 (0 or 3) so that the remaining sum (15−excluded) remains divisible by 3.

Set 1 (Exclude 0): {1,2,3,4,5}. Sum = 15. Arrangements = 5!=120.

Set 2 (Exclude 3): {0,1,2,4,5}. Sum = 12. 
Arrangements (excluding leading zeros) = 4 × 4! = 4 × 24 = 96.

Total: 120 + 96 = 216.

Correct Answer: E. 216

Explanation 2: Detailed Step-by-Step

To form a five-digit multiple of 3 from the digits {0,1,2,3,4,5}, we must follow these steps:

Step 1: Find valid groups of 5 digits

The sum of all six digits is 0 + 1 + 2 + 3 + 4 + 5 = 15. 
Since 15 is divisible by 3, if we remove one digit x, the remaining sum (15 − x) will only be divisible by 3 if x itself is a multiple of 3. The multiples of 3 in our set are 0 and 3.

Group A: Exclude 0. 
The digits are {1,2,3,4,5}. 
Sum = 15 (Divisible by 3).

Group B: Exclude 3. 
The digits are {0,1,2,4,5}. 
Sum = 12 (Divisible by 3).

Step 2: Calculate permutations for Group A {1,2,3,4,5} 
Since there is no zero in this set, any arrangement of these 5 digits will form a valid five-digit number.
Calculation: 5 × 4 × 3 × 2 × 1 = 5! = 120 ways.

Step 3: Calculate permutations for Group B {0,1,2,4,5} 
In this set, we have the digit 0. A five-digit number cannot start with 0.
First digit: Can be any of {1,2,4,5} (4 choices).
Remaining four digits: Can be arranged in the remaining 4 positions in 4! ways.
Calculation: 4 × (4 × 3 × 2 × 1) = 4 × 24 = 96 ways.

Step 4: Combine the results 
Total ways = (Ways from Group A) + (Ways from Group B) Total ways = 120 + 96 = 216.

Correct Answer: E. 216

TheAceClub · Answer

Answer: E

Explanation:
A number is divisible by 3 if the sum of its digits is a multiple of 3. You have six digits: {0, 1, 2, 3, 4, 5}. Their total sum is 15. To form a five-digit number without repetition, you must exclude exactly one digit. For the resulting five digits to sum to a multiple of 3, the excluded digit must itself be a multiple of 3.

Case 1: Exclude 0. Digits are {1, 2, 3, 4, 5}. Sum = 15.
Total arrangements = 5! = 120.

Case 2: Exclude 3. Digits are {0, 1, 2, 4, 5}. Sum = 12.
Total arrangements = 5! minus cases where 0 is in the first digit.
120 - 4! = 120 - 24 = 96.
Total = 120 + 96 = 216.

Trick:
Total sum of all available digits is 15. Since 15 is a multiple of 3, the digit you drop must be a multiple of 3 (0 or 3) to keep the remaining sum divisible by 3. This immediately narrows the problem to two scenarios instead of testing every combination.

Cheers,
The Ace Club by The Admit Co

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

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