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If N = 1000x + 100y + 10z, where x, y, and z are different positive integers less than 4, the remainder when N is divided by 9 is

(A) 2 (B) 4 (C) 6 (D) 8 (E) 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).

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.

Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]

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22 Oct 2012, 07:51

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Anichka wrote:

If N = 1000x + 100y + 10z, where x, y, and z are different positive integers less than 4, the remainder when N is divided by 9 is

(A) 2 (B) 4 (C) 6 (D) 8 (E) 9

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.
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PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]

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24 Oct 2012, 18:41

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mindmind wrote:

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

Since we have to divide N by 9, let's see N as = (999+1)x + (99+1)y + (9+1)z => N = 999x+99y+9z +x + y +z => N = 9(111x+11y+z) + (x + y + z)

We know that first part is always divisible by 9, since it is a multiple of 9. Remaining part is x+y+z; the remainder has to come from this part. But from the question, since these numbers are 1,2 and 3 (doesn't matter which is x or y or z), their sum is 6. So, remainder is 6.

Re: N = 1000x + 100y + 10z, remainder of N/9 [#permalink]

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06 Jun 2013, 07:41

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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

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