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x, y, and z are three consecutive multiples of 3 such that 0
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17 Jan 2020, 11:39
17 Jan 2020, 11:39
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x, y, and z are three consecutive multiples of 3 such that \( 0 < x < y < z \).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
The remainder when the sum of x + 1, y— 2, and z + 3 is divided by 9 |
2 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
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x, y, and z are three consecutive multiples of 3 such that 0
[#permalink]
19 Jan 2020, 07:03
19 Jan 2020, 07:03
1
Carcass wrote:
x, y, and z are three consecutive multiples of 3 such that \( 0 < x < y < z \).
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
The remainder when the sum of x + 1, y — 2, and z + 3 is divided by 9 |
2 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Here are some examples of consecutive multiples of 3:
6, 9, 12
24, 27, 30
51, 54, 57
Notice that each value is 3 greater than the value before it
So, if x, y, and z are three consecutive multiples of 3, then we can write:
x = x
y = x + 3
z = x + 6
Quantity A: The remainder when the sum of x + 1, y — 2, and z + 3 is divided by 9
We want the remainder when [(x + 1) + (y - 2) + (z + 3)] is divided by 9
Replace y and z with their equivalent values above to get:
[(x + 1) + (x + 3 - 2) + (x + 6 + 3)]
Simplify to get: 3x + 11
IMPORTANT: Since x is a multiple of 3, we know that x = 3k, for some integer k
So, we can take our sum of 3x + 11 and rewrite it as: 3(3k) + 11
Simplify to get: 9k + 11
We can rewrite this as: 9k + 9 + 2
And then factor to get: 9(k + 1) + 2
When we write it this way, we can see that are sum is TWO GREATER then some multiple of 9.
So when we divide the sum by 9, the remainder must be 2
Quantity B: 2
So, the two quantities are equal.
Answer: C
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x, y, and z are three consecutive multiples of 3 such that 0 [#permalink]
19 Jan 2020, 07:03
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