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?
If N = 1000x + 100y + 10z. . .
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A quick approach is to test a value of N that satisfies the given conditions.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
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
Cheers,
Brent
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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
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