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If N = 1000x + 100y + 10z. . .

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by Vincen » Sun Sep 24, 2017 6:22 am
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

The OA is C.

How could I find the remainder? Can I say that N is a the 4 digit number "xyz0" ? Or what are the calculus needed to solve this PS question?
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by Brent@GMATPrepNow » Sun Sep 24, 2017 6:46 am
Vincen 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
A quick approach is to test a value of N that satisfies the given conditions.

Given: x, y, and z are different positive integers less than 4
So, how about x = 1, y = 2 and z = 3
We get: N = 1000(1) + 100(2) + 10(3)
= 1000 + 200 + 30
= 1230

The remainder when N is divided by 9 is...
1230 divided by 9 = 136 with remainder 6

Answer: C

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Sun Sep 24, 2017 10:39 am
Hi Vincen,

Based on the 'limitations' provided by this question, N can only be one of 6 possible numbers:

1230
1320
2130
2310
3120
3210

The answer choices are 'fixed', which implies that one of those results will occur for ANY of those possible values for N. Thus, you can use any of the six options.

IF... N = 1230....
1230/9 = 136 r 6

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
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