If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
11
22
222
3
6
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Tough One.
This topic has expert replies
-
goyalsau
- Legendary Member
- Posts: 866
- Joined: Mon Aug 02, 2010 6:46 pm
- Location: Gwalior, India
- Thanked: 31 times
Saurabh Goyal
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
-
outreach
- Legendary Member
- Posts: 748
- Joined: Sun Jan 31, 2010 7:54 am
- Thanked: 46 times
- Followed by:3 members
is it 6?
considered the 3 digit no as 9,8,7..sum of 987,897,798 divisible by 6
tried another combination 1,2,3
considered the 3 digit no as 9,8,7..sum of 987,897,798 divisible by 6
tried another combination 1,2,3
-------------------------------------
--------------------------------------
General blog
https://amarnaik.wordpress.com
MBA blog
https://amarrnaik.blocked/
--------------------------------------
General blog
https://amarnaik.wordpress.com
MBA blog
https://amarrnaik.blocked/
-
goyalsau
- Legendary Member
- Posts: 866
- Joined: Mon Aug 02, 2010 6:46 pm
- Location: Gwalior, India
- Thanked: 31 times
OA is 222
Saurabh Goyal
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
-
shovan85
- Community Manager
- Posts: 991
- Joined: Thu Sep 23, 2010 6:19 am
- Location: Bangalore, India
- Thanked: 146 times
- Followed by:24 members
We can form a number in 6 ways as: abc,acb,bac,bca,cab,cba
Hence the actual 3 digit number will be 100a+10b+c, 100a+10c+b, 100b+10a+c, 100b+10c+a, 100c+10a+b, 100c+10b+a
Add all 6 of them we will get 222a+222b+222c = 222(a+b+c)
Thus it will be divisible by 222
Hence the actual 3 digit number will be 100a+10b+c, 100a+10c+b, 100b+10a+c, 100b+10c+a, 100c+10a+b, 100c+10b+a
Add all 6 of them we will get 222a+222b+222c = 222(a+b+c)
Thus it will be divisible by 222
-
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
The solutions offered above that arrived at the correct answer (222) are perfect, but for those who worry that they wouldn't be able to think that way during the test, another approach would be to plug in a few times.goyalsau wrote:If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
11
22
222
3
6
123+132+213+231+321+312 = 1332. 1332/222=6.
235+253+325+352+523+532 = 2220. 2220/222=10.
147+174+417+471+714+741 = 2664. 2664/222=12.
By now we could feel pretty secure that 222 is the largest factor that will work.
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
Hey, I solved this with a formula:
(a+b+c) (111) * (2P1) = (a+b+c) *111 * 2! = (a+b+c)*222
There fore the ans is 222
LOgic :
Because its a 3 digit no, multiply the sum of digits with 111 and because 3 digits are given, so at any point 1 digit is fixed you can arrange(permutation) 1 digit in 2 ways. So (2P1)
(a+b+c) (111) * (2P1) = (a+b+c) *111 * 2! = (a+b+c)*222
There fore the ans is 222
LOgic :
Because its a 3 digit no, multiply the sum of digits with 111 and because 3 digits are given, so at any point 1 digit is fixed you can arrange(permutation) 1 digit in 2 ways. So (2P1)
-
pesfunk
- Master | Next Rank: 500 Posts
- Posts: 286
- Joined: Tue Sep 21, 2010 5:36 pm
- Location: Kolkata, India
- Thanked: 11 times
- Followed by:5 members
Great trick...we can just plug values for 1 2 3 and check the largest divisor... Thanks Mitch 
GMATGuruNY wrote:The solutions offered above that arrived at the correct answer (222) are perfect, but for those who worry that they wouldn't be able to think that way during the test, another approach would be to plug in a few times.goyalsau wrote:If x represents the sum of all the positive three-digit numbers that can be constructed using each of the distinct nonzero digits a, b, and c exactly once, what is the largest integer by which x must be divisible?
11
22
222
3
6
123+132+213+231+321+312 = 1332. 1332/222=6.
235+253+325+352+523+532 = 2220. 2220/222=10.
147+174+417+471+714+741 = 2664. 2664/222=12.
By now we could feel pretty secure that 222 is the largest factor that will work.
