M04-32

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. 96
B. 120
C. 181
D. 200
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

rohitkumar77 · Answer

trick is to know which numbers are divisible by 3 , 
whose sum of the digits are divisible by 3.

with 1 , 2 ,3,4, and 5 we can make following numbers :-

5*4*3*2*1 = 120 numbers ;

here the sum is 15 for the digits of the numbers formed formed ;

next are the numbers whose sum of the digits is 12 .

taking 0 , 1 , 2 ,4 ,5 :- 
 we can make 96 numbers:- 4*4*3*2*1 = 96 ( 4 IN STARTING AS A NUMBER CANT START WITH ZERO )

TOTAL NUMBERS FORMED = 120+96 = 216

OPTION E THEREBY .

Barnal · Answer

I think this is a  high-quality  question and I agree with explanation. I think that this question is great.

matmen · Answer

I think this is a  high-quality  question and I agree with explanation.

AnirudhaS · Answer

So 5 digit number is a requirement without repeating.
There are in total 6 numbers - 0,1,2,3,4,5
Number is divisible by 3 means sum of the digits also divisible by 3
5+4+3+2+1=15 is divisible by 3
So total number of permutations = 5! =  120 
Now these number of combinations are without taking 0 into account.
If we take 0 into account, there is only one other choice of digits which is divisible by 3
This is 0,1,2,4,5
0+1+2+4+5 = 12 is also divisible by 3
Here we cannot take 0 as the first digit, otherwise it wont be a 5 digit number
Therefore possible permutations in this case = 4*4*3*2*1 =  96 
Total number of ways possible =  120 + 96  =  216

Answer:  E

malaka1 · Answer

My only question is why can't 0 be the left most digit? Is it just because it won't "form" a number?

Bunuel · Answer

We need  five -digit numbers. If 0 is the first digit, then we get a four-digit number. For example, 01234 is the same as 1234 and it's a four-digit number.

maravapalli · Answer

Can I divide total possible numbers = (600) by 3

kaladin123 · Answer

I think this is a  high-quality  question and I agree with the explanation.

AndrewN · Answer

Hello, everyone. I wanted to take a minute to explain why we  cannot  simply figure out the number of different five-digit (valid) combinations and then divide by 3. To put it simply, we are not dealing with a sequence of  successive  integers; rather, the five-digit integers will skip around (with some regularity). Consider just the first few combinations with the fixed stem 103-:

10324
10325

10342
10345

10352
10354

Are two of these integers divisible by 3, as we might expect? No. In fact, not a single one works, invoking the same divisibility property as that mentioned in the official solution. (The sum of the digits of each integer is 10, 11, 10, 13, 11, and 13, respectively.)

We have to read the question stem carefully and adapt our strategy accordingly. 200 might be a reasonable place to start as an estimate, but how could we then tell the difference between 181 and 216? This one is best approached a bit more technically (and trust me, I like to rely on logic for shortcuts).

Good luck with your studies.

- Andrew

juancarlos12 · Answer

First, we summarize the question:
 1) Identify valid cases: 
 A number abcde=3k ʌ the available numbers are: {0,1,2,3,4,5}
 So, we need to identify the valid numbers that follow this logic: a+b+c+d+e=3k
 Maximum sum = 15 (1+2+3+4+5)
 Minimum sum = 10 (0+1+2+3+4)

a)  With available numbers, the maximum sum is 15, so the 5 possible numbers are  {1,2,3,4,5} 
  b)  The sum can be 12=3k, so the 5 possible numbers are  {1,2,0,4,5} 
 c) The following sum can be 9, but remember that the Minimum sum is 10, so this is out!

2) Generate the permutations of the only 2 valid cases 
  a)   {1,2,3,4,5} 
 __ __ __ __ __
 5 x 4 x 3 x 2 x 1=120

b)   {1,2,0,4,5} 
 __ __ __ __ __
 4 x 4 x 3 x 2 x 1 =96 (The fist digit has only 4 posibilities, because it cannot be zero)

Finally, we add the 2 results, 120+96=216

Option E

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

aks0101 · Answer

I think this is a  high-quality  question and I agree with explanation.

Shivang29 · Answer

This is a great question that’s helpful for learning and I like the solution - it’s helpful.

siddharth_ · Answer

I like the solution - it’s helpful.

Sushi_545 · Answer

hi bunuel,

why 5*2*3*4*5

is not valid here?

first pos i five ways(except zero) * then the rest comn?

Bunuel · Answer

The logic of what you are doing is not clear. The short answer is that it is invalid because divisibility by 3 depends on the sum of the chosen five digits, not on their positions. Review the official solution carefully, as it shows that only certain digit sets with sums divisible by 3 are valid.

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