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If x, y, and z are consecutive integers such that x > y > z

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If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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New post Updated on: 21 Jul 2013, 09:19
10
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

59% (01:55) correct 41% (01:39) wrong based on 339 sessions

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If x, y, and z are consecutive integers such that x > y > z and the sum of x, y, and z is a multiple of 10, which of the following could be the value of x?

A. 0
B. 1
C. 9
D. 10
E. 12

Guys I think 2 answers are correct. What do you come up with?

Originally posted by vibhav on 21 Jul 2013, 08:55.
Last edited by Bunuel on 21 Jul 2013, 09:19, edited 1 time in total.
Edited the question.
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Re: If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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New post 21 Jul 2013, 09:19
vibhav wrote:
If x, y, and z are consecutive integers such that x > y > z and the sum of x, y, and z is a multiple of 10, which of the following could be the value of x?

A. 0
B. 1
C. 9
D. 10
E. 12

Guys I think 2 answers are correct. What do you come up with?


Given that \(z + y +x = (y-1) + y + (y+1) = 3y = 10k\) --> \(y\) is a multiple of 10 --> \(x = y+1\), so x is a multiple of 10 plus 1. Only answer choices B fits: \(z=-1\), \(y=0\), and \(z=1\) --> \(z+y+x=0\).

Answer: B.

Else you could simply plug-in values for x in (x-2) + (x-1) + x and see which one yields a multiple of 10.
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Re: If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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New post 21 Jul 2013, 09:53
Bunuel why can't it be 11,10,9 totals to 30 which is divisible by 10?
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Re: If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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New post 21 Jul 2013, 09:54
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Re: If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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New post 23 Dec 2014, 11:20
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Hi All,

This question can be easily beaten by a bit of "brute force" and TESTing THE ANSWERS.

We're given a very specific set of restrictions in this prompt:
1) X, Y and Z are CONSECUTIVE INTEGERS
2) X > Y > Z
3) X+Y+Z = a multiple of 10

We're asked which of the 5 answer COULD be the value of X given these restrictions. Rather than staring at the screen or doing layered math, we can "brute force" the answers until we find one that fits these restrictions..

Answer A: X = 0
In this case, the numbers would be 0, -1, and -2. The sum = -3 which is NOT a multiple of 10. Eliminate A.

Answer B: X = 1
In this case, the numbers would be 1, 0, and -1. The sum = 0 which IS a multiple of 10. B IS the answer.

Final Answer:

For the sake of argument, if you did not immediately realize that 0 is a multiple of 10, then you could quickly TEST the remaining 3 options and quickly disprove them.

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Re: If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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New post 25 Dec 2014, 09:20
1
I did it this way, which might be a bit similar in a way:

1) Since they are consecutive integers, these could be: x, x-1, x-2 (x>y>z)

2) I added them to give 0 (the 1st multiple of 10, if we say that they add up to 0*Unckown value):
x+x-1+x-2=0
3x-3=0
3x=3
x=3/3
x=1, Answer is B
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Re: If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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New post 25 Dec 2014, 10:42
Hi pacifist85,

Your approach works here, but ONLY because 0 is the multiple of 10 that is involved in the correct answer. If it was any other multiple of 10, then you would have only gotten the solution by "brute forcing" your equation into every possible multiple of 10.

eg.

3x - 3 = 10
3x - 3 = 20
3x - 3 = 30
Etc.

Since these answer choices in this question are all relatively small, even if you did have to brute force multiple possibilities, you would have gotten to the answer relatively quickly, so I think that your approach is fine. In the end, I measure any approach by 2 things:

1) Did it get you the correct answer.
2) Were you able to complete the question relatively quickly (for that prompt).

If the answer to both questions is YES, then you did well.

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Re: If x, y, and z are consecutive integers such that x > y > z  [#permalink]

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Re: If x, y, and z are consecutive integers such that x > y > z   [#permalink] 20 Aug 2018, 04:24
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