A five digits number divisible by 3 is to be formed using the numbers

Question

A five digits number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5, without repetition. What is the total number of ways this can be done?]

(A) 216 
(B) 240
(C) 600 
(D) 3125 
(D) 3325

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TarunKumar1234 · Accepted Answer

we have 6 digits and required nos. must be of 5 digits. It means, we can replace only 1 digit.

Sum of digits should be divisible by 3 then the no. will be divisible by 3.
Case1 : To choose from, 5, 4, 3, 2, 1 then 5! = 120 ways
Case 2: To choose from, 5, 4, 0, 2, 1 then 4*4! = 96 ways

So, Total nos. of ways = 120 +96=216

So, I think A.

Ramkumar2k · Answer

Here we need a five digit number divisible by 3 (so the sum of the five numbers should be divisible by 3)

So, it can be 0,1,2,4,5 or 1,2,3,4,5

So, it can be done by 5x4x3x2x1 ways for the two sets of numbers.

So, the answer is 2 x (5x4x3x2x1) = 240.

Option: B

gmatcafe · Answer

A number is divisible by 3 if sum of its digits is divisible by 3. From the given numbers we can form numbers that are divisible by 3 in 2 ways:
1) From the numbers 1,2,3,4, and 5: 5!=120 different numbers can be formed without repetition. 
2) From the numbers 0,1,2,4 and 5: 5!-4!=120-24=96 (we subtract 4! from 5! because 0 can not be the first digit) different numbers can be formed without repetition.
In total 120+96=216 different numbers that are divisible by 3 can be formed from the given numbers without repetition.
IMO Answer is A)

ScottTargetTestPrep · Answer

Solution:

Recall that if the sum of the digits of a number is divisible by 3, then it is divisible by 3. We see that the sum of digits 0 to 5, inclusive, is 15, which is divisible by 3. However, there are 6 digits; so if we need to choose 5 of them and still want the sum of the digits to be divisible by 3, we must exclude a digit that is divisible by 3. That is, we must exclude either 0 or 3.

Case 1: Excluding 0

If we exclude 0, we will be using the digits 1 to 5, inclusive, to form a 5-digit number. Notice that the sum of these digits is 15, so no matter how we arrange these digits to form a 5-digit number, it will be divisible by 3. Therefore, there are 5! = 120 such numbers in this case.

Case 2: Excluding 3

If we exclude 3, we will be using the digits 0, 1, 2, 4, and 5 to form a 5-digit number. Notice that the sum of these digits is 12, so no matter how we arrange these digits to form a 5-digit number (assuming the digit 0 is allowed to be the first digit of the number), it will be divisible by 3. Therefore, there will also be 5! = 120 such numbers in this case if we permit the digit 0 as the first digit of the number. However, since we can’t have the digit 0 as the first digit of a number, we have to exclude one-fifth, or 24, of the 120 numbers. Therefore, the actual number of numbers in this case is 120 - 24 = 96.

Therefore, there are a total of 120 + 96 = 216 such numbers.

Answer: A

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A five digits number divisible by 3 is to be formed using the numbers

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