N = 1000x + 100y + 10z, remainder of N/9

Since we cannot have more than one correct answers in PS questions, then pick some numbers for x, y, and z and find the reminder when 1000x + 100y + 10z is divided by 9.

Say x=1, y=2, and z=3, then 1000x + 100y + 10z = 1,230 --> 1,230 divide by 9 yields the remainder of 6 (1,224 is divisible by 9 since the sum of its digit is a multiple of 9, thus 1,230, which is 6 more than a multiple of 9, yields the remainder of 6 when divided by 9).

Answer; C.

Hope it's clear.

Is there any other method to solve the same.... instead of plugging in the values and dividing them by 9..

Hi Bunuel - But what happens when x = 2, y =3 and z = 1. Then the reminder is something different. Can we expect such questions in GMAT. Thanks

Since we cannot have more than one correct answer in PS questions, then for ANY values of x, y, and z which satisfy the given condition (different positive integers less than 4) we must get the same answer, otherwise the question would be flawed.

If you check the remainder when x = 2, y =3 and z = 1, then you'll see that it'll still be 6.

Hope it's clear.

Since x, y, and z are distinct positive integers less than 4, they can be 1, 2, or 3, doesn't matter which one is what.
Then N = 10(100x + 10y + z) = xyz0 is 10 times a three digit number formed by any of the permutations of the digits 1, 2, and 3.
For example 1230, 3210, 3120,...in fact, is quite a short list, only 3! = 6 numbers.
The divisibility rule by 9 says that the sum of the digits of the number and the number itself, leave the same remainder when divided by 9.
For example, when the remainder is 0, the number and the sum of its digit, are both divisible by 9.

Since the sum of the digits of N is 1 + 2 + 3 = 6, the remainder when N is divided by 9, is 6.

Answer C.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

For any number, it is divisible by 9 if all of the individual numbers in the number are divisible by 9 when added together.
For example:
18 (9x2): 1+8 = 9/9 = 1,
36 (9x4): 3+6 = 9/9 = 1
396 (9x44): 3+6+9 = 18/9 = 2
657 (9x73): 6+5+7 = 18/9 = 2
999 (9x111): 9+9+9 = 27/9 = 3
etc.

The remainder when the individual numbers in a number are divided by 9 is the remainder when the number is divided by 9 as well
For example:
19 (9x2 r1): 1+9 = 10/9 = 1 r1
38 (9x4 r2): 3+8 =11/9 = 1 r2
397 (9x44 r1): 3+9+7 = 19/9 = 2 r1
659 (9x73 r2): 6+5+9 = 20/9 = 2 r2
1002 (9x111 r3): 1+0+0+2 = 3/9 = 0 r3
etc.

That being said, in the problem if x, y, and z are all different positive integers less than 4, they can be 1,2, and 3 in any order:
1230, 1320, 2130, 2310, 3120, and 3210
if we apply the rule above:
1230: 1+2+3+0 = 6/9 = 0 r6
1320: 1+3+2+0 = 6/9 = 0 r6
2130: 2+1+3+0 = 6/9 = 0 r6
etc. etc. etc.

If you know this rule, then all you would need to know is x+y+z = 1+2+3 = 6

Answer is C

To start, Let x = 1; y = 2 and z = 3.

=> N = 1000 + 200 + 30 = 1230 = 1 + 2 + 3 + 0 = 6

=> N = 6, when divided by 9, will give remainder as 6

If x = 3; y = 2 and z = 1.

=> N = 3000 + 200 + 10 = 3210 = 3 + 2 + 1 + 0 = 6

=> N = 6, when divided by 9, will give remainder as 6


Hence, the digits will remain the same. Therefore, the remainder will be 6.

Answer C