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by goyalsau » Mon Oct 04, 2010 8:24 am
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
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by outreach » Mon Oct 04, 2010 8:40 am
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
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by goyalsau » Mon Oct 04, 2010 8:53 am
OA is 222
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by selango » Mon Oct 04, 2010 9:08 am
X=(100a+10b+c)+(100a+10c+b)+(100b+10a+c)+(100b+10c+a)+(100c+10a+b)+(100c+10b+a)

X=222a+222b+222c=222(a+b+c)

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by shovan85 » Mon Oct 04, 2010 9:09 am
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
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by GMATGuruNY » Mon Oct 04, 2010 1:03 pm
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
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.

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.
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by himanshi » Sun Nov 07, 2010 1:37 pm
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)
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by pesfunk » Sun Nov 07, 2010 6:22 pm
Great trick...we can just plug values for 1 2 3 and check the largest divisor... Thanks Mitch :D
GMATGuruNY wrote:
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
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.

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.
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