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How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number m

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by BTGModeratorVI » Wed Jul 29, 2020 2:51 pm

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How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144

Answer: E
Source: Veritas Prep
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Re: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the numb

by Scott@TargetTestPrep » Sat Aug 01, 2020 6:41 am
BTGModeratorVI wrote: ↑
Wed Jul 29, 2020 2:51 pm
 How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144

Answer: E
Solution:

Recall that if a number is divisible by 4, the last two digits of the number must be divisible by 4. Therefore, the last two digits of the 5-digit number must be 04, 12, 20, 24, 32, 40 or 52. Furthermore, since the first digit can’t be 0, we can divide the last two digits into two groups: 1) 04, 20, 40 (where the digit 0 is used) and 2) 12, 24, 32, 52 (where the digit 0 is not used).

Group 1: The digit 0 is used as one of the last two digits

Since two digits (including 0) are used, there are 4 options for the first digit, 3 options for the second digit and 2 options for the third digit. Since the last two digits together have 3 options, the total number of five-digit numbers can be formed is 4 x 3 x 2 x 3 = 72.

Group 2: The digit 0 is not used as one of the last two digits

Since two digits (excluding 0) are used, there are 3 options for the first digit (since it can’t be 0), 3 options for the second digit and 2 options for the third digit. Since the last two digits together have 4 options, the total number of five-digit numbers can be formed is 3 x 3 x 2 x 4 = 72.

Therefore, there are a total of 72 + 72 = 144 five-digit numbers can be formed.

Answer: E

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Re: How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the numb

by swerve » Sat Aug 01, 2020 7:27 am
BTGModeratorVI wrote: ↑
Wed Jul 29, 2020 2:51 pm
 How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

A. 36
B. 48
C. 72
D. 96
E. 144

Answer: E
Source: Veritas Prep
You can try as follows,

Total of \(600\) possible number: \(5\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\) (zero can't be the \(1\)st digit)

Of those \(600\) number, \(300\) are even, because you have \(3\) odd and \(3\) even numbers. Of those \(300\) even numbers, about half are multiple of \(4\).

So, the correct answer is E
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