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If a, b, and c are multiples of 3 such that a > b > c > 0, w
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12 Aug 2018, 16:08
12 Aug 2018, 16:08
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Question Stats:
85% (00:55) correct
14% (00:39) wrong based on 92 sessions
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If a, b, and c are multiples of 3 such that a > b > c > 0, which of the following values must be divisible by 3?
Indicate all such values.
A. a + b + c
B. a – b + c
C. \(\frac{abc}{9}\)
Indicate all such values.
A. a + b + c
B. a – b + c
C. \(\frac{abc}{9}\)
Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
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17 Aug 2018, 16:00
17 Aug 2018, 16:00
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Expert Reply
Explanation
Since a, b, and c are all multiples of 3, a = 3x, b = 3y, c = 3z, where x > y > z > 0 and all are integers.
Substitute these new expressions into the statements.
First statement: \(a + b + c = 3x + 3y + 3z = 3(x + y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.
Second statement: \(a - b + c = 3x - 3y + 3z = 3(x - y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.
Third statement:\(\frac{abc}{9}=\frac{3x \times 3y \times 3z}{9}=\frac{27xyz}{9}= 3xyz\). Since xyz is an integer, this number must be divisible by 3.
Since a, b, and c are all multiples of 3, a = 3x, b = 3y, c = 3z, where x > y > z > 0 and all are integers.
Substitute these new expressions into the statements.
First statement: \(a + b + c = 3x + 3y + 3z = 3(x + y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.
Second statement: \(a - b + c = 3x - 3y + 3z = 3(x - y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.
Third statement:\(\frac{abc}{9}=\frac{3x \times 3y \times 3z}{9}=\frac{27xyz}{9}= 3xyz\). Since xyz is an integer, this number must be divisible by 3.
Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
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06 Jul 2019, 13:36
06 Jul 2019, 13:36
sandy wrote:
If a, b, and c are multiples of 3 such that a > b > c > 0, which of the following values must be divisible by 3?
Indicate all such values.
A. a + b + c
B. a – b + c
C. \(\frac{abc}{9}\)
Indicate all such values.
A. a + b + c
B. a – b + c
C. \(\frac{abc}{9}\)
There is a little bit problem with this type of question, like when you solve the question. and get two or three answer choices, and when you just put your answer in one case (A), then try to put the second answer in (B/C) but before this, the GRE Timer just stop when you click answer A. Need some Web coding to dynamic this. Thanks
