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

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

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) Pick C

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

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.

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)

Great trick...we can just plug values for 1 2 3 and check the largest divisor... Thanks Mitch 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.