If x, y, z are integers, is xyz a multiple of 6?

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Fri Sep 21, 2018 12:16 am

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[Math Revolution GMAT math practice question]

If x, y, z are integers, is xyz a multiple of 6?

1) x+y+z is a multiple of 6
2) x, y, and z are consecutive integers

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Source: Beat The GMAT — Data Sufficiency | Fri Sep 21, 2018 12:16 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

If x, y, z are integers, is xyz a multiple of 6?

by fskilnik@GMATH » Fri Sep 21, 2018 7:06 am

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x, y, z are integers, is xyz a multiple of 6?

1) x+y+z is a multiple of 6
2) x, y, and z are consecutive integers

\[x,y,z\,\,\,{\text{ints}}\]
\[\frac{{xyz}}{6}\,\,\mathop = \limits^? \,\,\operatorname{int} \]
\[\left( 1 \right)\,\,\,\frac{{x + y + z}}{6} = \operatorname{int} \,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,y,z} \right) = \left( {0,0,0} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,\left( {x,y,z} \right) = \left( {2,2,2} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,\left\{ \begin{gathered}
\hfill \left( {{\text{at}}\,\,{\text{least}}} \right)\,\,{\text{one}}\,\,{\text{even}} \\
\hfill {\text{exactly}}\,\,{\text{one}}\,\,{\text{multiple}}\,\,{\text{of}}\,\,3 \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]

The correct answer is therefore [spoiler]__(B)____ [/spoiler] .

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Sat Sep 22, 2018 8:42 pm

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x, y, z are integers, is xyz a multiple of 6?

1) x+y+z is a multiple of 6
2) x, y, and z are consecutive integers

Given: x, y, z are integers
Question: Is xyz a multiple of 6?

Let's take each statement one by one.

1) x + y + z is a multiple of 6.

Case 1: Say x, y and z are 1, 2, and 3, respectively, then xyz = 1*2*3 = 6. The answer is yes.
Case 2: Say x = y = 1 and z = 4, respectively, then xyz = 1*1*4 = 4. The answer is No. No unqiue answer.

2) x, y, and z are consecutive integers.

Note that three consecutive integers guarantee that there is at least one even number (this assures that there is at least one multiple of 2) and at least one multiple of 3.

For any number to be a multiple of 6, it must be a multiple of 2 as well as 3; thus, Statement 2 suffices.

Take a couple of examples:

1. Say the integers are 1, 2 and 3; we have one even and a multiple of 3
2. Say the integers are 11, 12 and 13; we have one even and a multiple of 3

The correct answer: B

Hope this helps!

-Jay
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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Mon Sep 24, 2018 4:07 am

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since x and y are two consecutive integers, one of them, x or y is an even integer, and xy is a multiple of 2.
Since x, y and z are three consecutive integers, one of them, x, y or z is a multiple of 3, and xyz is a multiple of 3.
Therefore, xyz is a multiple of 6.
Both conditions (together) are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If x = 1, y = 2 and z = 3, then xyz = 6 is not a multiple of 6.
If x = 2, y = 2 and z = 2, then xyz = 8 is a multiple of 8.
Since we do not have a unique answer, condition 1) is not sufficient.

Condition 2)
Using the same argument as the one given for both conditions together, condition 2) is sufficient.

Therefore, B is the answer.
Answer: B

In cases where 3 or more additional equations are required, such as for original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since x and y are two consecutive integers, one of them, x or y is an even integer, and xy is a multiple of 2.
Since x, y and z are three consecutive integers, one of them, x, y or z is a multiple of 3, and xyz is a multiple of 3.
Therefore, xyz is a multiple of 6.
Both conditions (together) are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If x = 1, y = 2 and z = 3, then xyz = 6 is not a multiple of 6.
If x = 2, y = 2 and z = 2, then xyz = 8 is a multiple of 8.
Since we do not have a unique answer, condition 1) is not sufficient.

Condition 2)
Using the same argument as the one given for both conditions together, condition 2) is sufficient.

Therefore, B is the answer.
Answer: B

In cases where 3 or more additional equations are required, such as for original conditions with "3 variables", or "4 variables and 1 equation", or "5 variables and 2 equations", conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

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