VJesus12 wrote: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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
