Veritas Test2 Q23

Hi Abhijit,

I approached this problem with simple counting.

Divisibilty rule for 4: Last two numbers must be divisible by 4

possibilities:
04 - 4*3*2
12 - 3*3*2
20 - 4*3*2
24 - 3*3*2
32 - 3*3*2
40 - 4*3*2
52 - 3*3*2

Count - 4 * 3 * 3 * 2 * 2 = 144

Answer E

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

For the 5-digit integer to be a multiple of 4, the last two digits must form a multiple of 4.
Options:
04, 12, 20, 24, 32, 40, 52.

Case 1: Last 2 digits include 0
Number options for the last 2 digits = 3. (04, 20 or 40.)
Number of options for the ten-thousands place = 4. (Any of the 4 remaining digits.)
Number of options for the thousands place = 3. (Any of the 3 remaining digits.)
Number of options for the hundreds place = 2. (Either of the 2 remaining digits.)
To combine these options, we multiply:
3*4*3*2 = 72.

Case 2: Last 2 digits do NOT include 0
Number options for the last 2 digits = 4. (12, 24, 32, or 52.)
Number of options for the ten-thousands place = 3. (Any of the remaining 4 digits but 0.)
Number of options for the thousands place = 3. (Any of the 3 remaining digits.)
Number of options for the hundreds place = 2. (Either of the 2 remaining digits.)
To combine these options, we multiply:
4*3*3*2 = 72.

Total options = 72+72 = 144.

The correct answer is E.