Hi All,
Can anyone explain how this one will be solved? Answer and explanation hidden here, but I don't get the logic.
How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?
A. 15
B. 90
C. 216
D. 120
E. 625
Answer [spoiler]C
And this is the explanation I found:
Test of divisibility for 3 : The sum of the digits of any number that is divisible by '3' is divisible by 3.
There are six digits, 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.
The sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits are divisible by '3'.
Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either '0' or '3' while forming the five digit numbers.
Case 1
If we do not use '0', then the remaining 5 digits can be arranged in 5! ways = 120 numbers.
Case 2
If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'. The first digit from the left can be any of the 4 digits 1, 2, 4 or 5. Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.
So, there will be 4*4! numbers = 4*24 = 96 numbers.
Combining Case 1 and Case 2, there are a total of 120 + 96 = 216 five digit numbers divisible by '3' that can be formed using the digits 0 to 5.[/spoiler]
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If the sum of the digits of integer N is a multiple of 3, then N itself is a multiple of 3.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?
A. 15
B. 96
C. 216
D. 120
E. 625
Adding 5 of the digits above, there are 2 ways to get a sum that is a multiple of 3 if no digit is repeated:
1+2+3+4+5 = 15 and 0+1+2+4+5 = 12.
Number of ways to arrange 1,2,3,4,5 = 5! = 120.
Number of 5-digit integers composed of 0,1,2,4,5:
Ten-thousands digit can be 1,2,4,5 = 4 choices.
Number of ways to arrange the remaining 4 digits = 4! = 24.
Combining our choices for the digits, we get:
Number of possible integers = 4*24 = 96.
Thus, total possible integers = 120+96 = 216.
The correct answer is C.
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Our number's digits must sum to 3. Since
0 + 1 + 2 + 3 + 4 + 5 = 15
we can only remove a digit that is ITSELF divisible by 3, e.g. 15 - 0 = 15 and 15 - 3 = 12. So our two sets of digits are
{1, 2, 3, 4, 5}
and
{0, 1, 2, 4, 5}
In the first case we have 5! options, or 120. In the second, we can't lead with 0, so we have 4 options for the leading digit, then 4! for the rest, for a total of 4! * 4 or 96 options.
120 + 96 => 216, so we're set! (Bad pun ...)
0 + 1 + 2 + 3 + 4 + 5 = 15
we can only remove a digit that is ITSELF divisible by 3, e.g. 15 - 0 = 15 and 15 - 3 = 12. So our two sets of digits are
{1, 2, 3, 4, 5}
and
{0, 1, 2, 4, 5}
In the first case we have 5! options, or 120. In the second, we can't lead with 0, so we have 4 options for the leading digit, then 4! for the rest, for a total of 4! * 4 or 96 options.
120 + 96 => 216, so we're set! (Bad pun ...)
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If a number is divisible by 3, the SUM of the digits will be divisible by 3.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?
A. 15
B. 96
C. 216
D. 120
E. 625
0 + 1 + 2 + 3 + 4 + 5 = 15 (which is divisible by 3)
We need to remove 1 digit (to create a 5-digit number). So, to ensure that the SUM of the remaining 5 digits is divisible by 3, the digit that we remove must be divisible by 3.
That means, we can remove EITHER 0 or 3
Removing 0 leaves us with the digits 1, 2, 3, 4, and 5, which have a sum of 15. Great.
Removing 3 leaves us with the digits 0, 1, 2, 4, and 5, which have a sum of 12. Great.
So, how many 5-digit numbers can we create with the digits 1, 2, 3, 4, and 5, and how many 5-digit numbers can we create with the digits 0, 1, 2, 4, and 5?
Start with the digits 1, 2, 3, 4, and 5
We cannot repeat digits.
So, we have 5 options for the first digit in the number.
We have 4 options for the second digit in the number.
We have 3 options for the third digit in the number.
We have 2 options for the fourth digit in the number.
We have 1 option for the fifth digit in the number.
So, the TOTAL number of 5-digit numbers = (5)(4)(3)(2)(1) = 120
NOTE: we haven't yet counted all of the 5-digit numbers can we create with the digits 0, 1, 2, 4, and 5
This means our final answer must be GREATER than 120.
So, we can ELIMINATE answer choices A, B, and D
IMPORTANT: IF we were to start listing 5-digit numbers that can be created with the digits 0, 1, 2, 4, and 5, we would have to ensure that the first digit is NOT 0. Otherwise, we'd get a 4-digit number (e.g., 02451 is NOT a 5-digit number).
This means that our list of 5-digit numbers (using 0, 1, 2, 4, and 5) will have FEWER THAN 120 numbers.
This means we can ELIMINATE answer choice E.
This leaves only answer choice C
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Brent
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