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
The OA is the option E.
How can I find the correct answer is a short way? I don't want to make the list <i class="em em-cry"></i> Help me!!!
Source: Veritas Prep
Target Test Prep is 20% off right now!
Use code FLASH20
Available with Beat the GMAT members only code
MORE DETAILS
How many five-digit numbers can be formed from the digits
This topic has expert replies
-
GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
For the 5-digit integer to be a multiple of 4, the last two digits must form a multiple of 4.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
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
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
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7263
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
To be divisible by 4, the last two digits of the number must be divisible by 4. Therefore, they can be 04, 12, 20, 24, 32, 40, and 52. We can split these into two groups: 1) 04, 20, 40, and 2) 12, 24, 32, 52VJesus12 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
The OA is the option E.
Group 1:
If the last two digits are 04, then there are 4 choices for the first (or ten-thousands) digit, 3 choices for the second (or thousands) digit, and 2 choices for the third (or hundreds) digit. So we have 4 x 3 x 2 = 24 such numbers if the last two digits are 04. Also there should be 24 numbers if the last two digits are 20 or 40. So we have 24 x 3 = 72 numbers in this group.
Group 2:
If the last two digits are 12, then there are 3 choices for the first (or ten-thousands) digit (since it can't be 0), 3 choices for the second (or thousands) digit, and 2 choices for the third (or hundreds) digit. So we have 3 x 3 x 2 = 18 such numbers if the last two digits are 12. Also there should be 18 numbers if the last two digits are 24, 32 or 52. So we have 18 x 4 = 72 numbers in this group also.
Therefore, there are a total of 72 + 72 = 144 numbers.
Answer: E
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
