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If x and y are positive integer and xy is divisible by 4 [#permalink]
05 Nov 2012, 17:18

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Question Stats:

75% (01:00) correct
25% (00:57) wrong based on 100 sessions

If \(x\) and \(y\) are positive integer and \(xy\) is divisible by 4, which of the following must be true?

A) If \(x\) is even then \(y\) is odd. B) If \(x\) is odd then \(y\) is a multiple of 4. C) If \(x+y\) is odd then \(y/x\) is not an integer. D) If \(x+y\) is even then \(x/y\) is an integer. E) \(x^y\) is even.

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
05 Nov 2012, 18:14

1

This post received KUDOS

Expert's post

gmatchase wrote:

If \(x\) and \(y\) are positive integer and \(xy\) is divisible by 4, which of the following must be true? A) If \(x\) is even then \(y\) is odd. B) If \(x\) is odd then \(y\) is a multiple of 4. C) If \(x+y\) is odd then \(y/x\) is not an integer. D) If \(x+y\) is even then \(x/y\) is an integer. E) \(x^y\) is even.

Please explain with your answer

I'm happy to help with this.

First of all, this is a very challenging GMAT problem --- it really demands a great deal of number sense to answer this efficiently.

The prompt tells us x & y are positive integer (mercifully!) and x*y is divisible by 4. That could happen if either one of them is a multiple of four, or if both of them are even.

(A) Counter-example: If x = 2, then y could not be odd, because 2*(odd) will never be divisible by four. NOT ALWAYS TRUE. (B) If x is odd, then there are no factors of 2 in x. There are two factors of 2 in x*y, since x*y is be divisible by 4. If those two factors of 2 don't come from x, they must come from y. Therefore y MUST BE divisible by four. (C) Counter-example: x = 3 and y = 12 --- x + 12 = 15 is odd, but y/x = 12/3 = 4 is an integer. NOT ALWAYS TRUE. (D) Counter-example: x = 12 and y = 8 ---- x + y = 20 is even, but x/y = 12/8 = 3/2 is not an integer. NOT ALWAYS TRUE. (E) Counter-example: x = 3 and y = 4 ---- x^y = 3^4 = 81 is not even. NOT ALWAYS TRUE.

In every pair in the counter-examples, the numbers were chosen so that x*y is divisible by 4, and any additional condition in that particular choice is met.

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
06 Nov 2012, 02:39

Expert's post

gmatchase wrote:

Quote:

If \(x\) and \(y\) are positive integer and \(xy\) is divisible by 4, which of the following must be true?

Thank you for the reply Mike. When I saw \(xy\), I didn't think that it is \(x*y\). I thought it might be below

\(x\) --> 1 \(y\) --> 4

then \(xy\) is 14. Is this not the case? If not, can you explain when the expression indicates \(x*y\)?

If xy meant to be two-digit integer xy, then it would be mentioned something like "x and y are digits of two-digit integer xy".

If \(x\) and \(y\) are positive integer and \(xy\) is divisible by 4, which of the following must be true?

A. If \(x\) is even then \(y\) is odd. B. If \(x\) is odd then \(y\) is a multiple of 4. C. If \(x+y\) is odd then \(\frac{y}{x}\) is not an integer. D. If \(x+y\) is even then \(\frac{x}{y}\) is an integer. E. \(x^y\) is even.

Notice that the question asks which of the following MUST be true not COULD be true.

A. If \(x\) is even then \(y\) is odd --> not necessarily true, consider: \(x=y=2=even\);

B. If \(x\) is odd then \(y\) is a multiple of 4 --> always true: if \(x=odd\) then in order \(xy\) to be a multiple of 4 y mst be a multiple of 4;

C. If \(x+y\) is odd then \(\frac{y}{x}\) is not an integer --> not necessarily true, consider: \(x=1\) and \(y=4\);

D. If \(x+y\) is even then \(\frac{x}{y}\) is an integer --> not necessarily true, consider: \(x=2\) and \(y=4\);

E. \(x^y\) is even --> not necessarily true, consider: \(x=1\) and \(y=4\);

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
06 Nov 2012, 09:54

1

This post received KUDOS

Expert's post

gmatchase wrote:

Thank you for the reply Mike. When I saw xy, I didn't think that it is x*y. I thought it might be below

x --> 1 y --> 4

then xy is 14. Is this not the case? If not, can you explain when the expression indicates x*y?

Dear gmatchase

First of all, I'd like to suggest --- don't over-do the "math" notation. That's useful for fractions & exponents, but for ordinary x+y, we don't need the special notation.

The blanket assumption throughout all of algebra is that writing two variables next to each other (mx) or writing a number directly next to a variable (5y) always means multiplication. That's the meaning 100% of the time, unless something different is very explicitly specified. If the problem wanted the variables to represent digits, first of all, it probably would have used letters from the beginning of the alphabet, certainly not x & y (the standard algebraic variables), and it would have to say, quite explicitly, something like "Let a and b be digits in a the two-digit number ab." In other words, because the blank assumption is --- two variables next to each other always means multiply --- when a problem wants to indicate digits or anything else, it practically has to set up traffic cones and flashing signs to indicate the change. For a digits problem, it would have to explain, explicitly, what each variable represented. Any problem that just tosses an xy at you with no verbiage will never be using those variable as digits --- that xy will be multiplication 100% of the time.

Here's an example of a rare digit problem. http://gmat.magoosh.com/questions/54 When you submit you answer, the following page will have a full video explanation.

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
06 Nov 2012, 15:18

mikemcgarry wrote:

gmatchase wrote:

Thank you for the reply Mike. When I saw xy, I didn't think that it is x*y. I thought it might be below

x --> 1 y --> 4

then xy is 14. Is this not the case? If not, can you explain when the expression indicates x*y?

Dear gmatchase

First of all, I'd like to suggest --- don't over-do the "math" notation. That's useful for fractions & exponents, but for ordinary x+y, we don't need the special notation.

The blanket assumption throughout all of algebra is that writing two variables next to each other (mx) or writing a number directly next to a variable (5y) always means multiplication. That's the meaning 100% of the time, unless something different is very explicitly specified. If the problem wanted the variables to represent digits, first of all, it probably would have used letters from the beginning of the alphabet, certainly not x & y (the standard algebraic variables), and it would have to say, quite explicitly, something like "Let a and b be digits in a the two-digit number ab." In other words, because the blank assumption is --- two variables next to each other always means multiply --- when a problem wants to indicate digits or anything else, it practically has to set up traffic cones and flashing signs to indicate the change. For a digits problem, it would have to explain, explicitly, what each variable represented. Any problem that just tosses an xy at you with no verbiage will never be using those variable as digits --- that xy will be multiplication 100% of the time.

Here's an example of a rare digit problem. http://gmat.magoosh.com/questions/54 When you submit you answer, the following page will have a full video explanation.

Does all this make sense?

Mike

Thank you again for all your explanation. It makes sense now.

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
25 Jun 2013, 08:23

Bunuel wrote:

gmatchase wrote:

Quote:

If \(x\) and \(y\) are positive integer and \(xy\) is divisible by 4, which of the following must be true?

Thank you for the reply Mike. When I saw \(xy\), I didn't think that it is \(x*y\). I thought it might be below

\(x\) --> 1 \(y\) --> 4

then \(xy\) is 14. Is this not the case? If not, can you explain when the expression indicates \(x*y\)?

If xy meant to be two-digit integer xy, then it would be mentioned something like "x and y are digits of two-digit integer xy".

If \(x\) and \(y\) are positive integer and \(xy\) is divisible by 4, which of the following must be true?

A. If \(x\) is even then \(y\) is odd. B. If \(x\) is odd then \(y\) is a multiple of 4. C. If \(x+y\) is odd then \(\frac{y}{x}\) is not an integer. D. If \(x+y\) is even then \(\frac{x}{y}\) is an integer. E. \(x^y\) is even.

Notice that the question asks which of the following MUST be true not COULD be true.

A. If \(x\) is even then \(y\) is odd --> not necessarily true, consider: \(x=y=2=even\);

B. If \(x\) is odd then \(y\) is a multiple of 4 --> always true: if \(x=odd\) then in order \(xy\) to be a multiple of 4 y mst be a multiple of 4;

C. If \(x+y\) is odd then \(\frac{y}{x}\) is not an integer --> not necessarily true, consider: \(x=1\) and \(y=4\);

D. If \(x+y\) is even then \(\frac{x}{y}\) is an integer --> not necessarily true, consider: \(x=2\) and \(y=4\);

E. \(x^y\) is even --> not necessarily true, consider: \(x=1\) and \(y=4\);

Answer: B.

What if xy= 44 ( x=11, y= 4 or x=4 and y= 11)

Option B, C and D hold true then....i know in must be true questions picking numbers can be confusing..

EDIT- (A) Counter-example: If x = 2, then y could not be odd, because 2*(odd) will never be divisible by four. NOT ALWAYS TRUE. (B) If x is odd, then there are no factors of 2 in x. There are two factors of 2 in x*y, since x*y is be divisible by 4. If those two factors of 2 don't come from x, they must come from y. Therefore y MUST BE divisible by four. (C) Counter-example: x = 3 and y = 12 --- x + 12 = 15 is odd, but y/x = 12/3 = 4 is an integer. NOT ALWAYS TRUE. (D) Counter-example: x = 12 and y = 8 ---- x + y = 20 is even, but x/y = 12/8 = 3/2 is not an integer. NOT ALWAYS TRUE. (E) Counter-example: x = 3 and y = 4 ---- x^y = 3^4 = 81 is not even. NOT ALWAYS TRUE. _________________

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
25 Jun 2013, 09:20

1

This post received KUDOS

Expert's post

prateekbhatt wrote:

What if xy= 44 ( x=11, y= 4 or x=4 and y= 11)

Option B, C and D hold true then....i know in must be true questions picking numbers can be confusing.

Dear prateekbhatt Yes. Remember that in any "must be true" math question, one answer has to be true, and most of the time, the other four will be usually be true --- some will be things that seem like they should be true. The GMAT simply is not going to give you one 100% true choice and four 100% false choices ---- it's going to be something more like one 100% true choice and four 90-95% true choices.

This means that, for any pair of numbers you pick, unless you are exceedingly lucky (or exceedingly skillful at picking numbers), more than one answer choice will still be true. The value of any single pair of numbers lies in [b]the choices you can eliminate[/b] ---- you pick one pair, eliminate some choices, then pick another pair, eliminate more choices, and whittle down the options. You are in the wrong frame of mind if you are trying to find the idea choice so that you can eliminate the four incorrect options all at once. If that happens by chance, great, but don't spend any time striving for that. Instead, think simplicity and speed. If I were plugging in numbers for this, my first choices would be the simplest kindergarten possibilities --- x = 1 & y = 4, then x = 4 & y = 1, then x = 2 & y = 2 --- in the first 10-20 seconds, you should be able to eliminate at least three answer choices with very easy selections for the variable values. Focus speed & efficiency, not on making ideal choices. Once you are down to two answers, you may need a more sophisticated choice --- for example, none of those choices would eliminate (D), so you would have to choose something like x = 2 & y = 8. That eliminates everything except the OA, (B), and I didn't use anything other than single-digit choices. Often you can remain with single digit math and get everything you need.

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
25 Jun 2013, 21:27

mikemcgarry wrote:

prateekbhatt wrote:

What if xy= 44 ( x=11, y= 4 or x=4 and y= 11)

Option B, C and D hold true then....i know in must be true questions picking numbers can be confusing.

Dear prateekbhatt Yes. Remember that in any "must be true" math question, one answer has to be true, and most of the time, the other four will be usually be true --- some will be things that seem like they should be true. The GMAT simply is not going to give you one 100% true choice and four 100% false choices ---- it's going to be something more like one 100% true choice and four 90-95% true choices.

This means that, for any pair of numbers you pick, unless you are exceedingly lucky (or exceedingly skillful at picking numbers), more than one answer choice will still be true. The value of any single pair of numbers lies in [b]the choices you can eliminate[/b] ---- you pick one pair, eliminate some choices, then pick another pair, eliminate more choices, and whittle down the options. You are in the wrong frame of mind if you are trying to find the idea choice so that you can eliminate the four incorrect options all at once. If that happens by chance, great, but don't spend any time striving for that. Instead, think simplicity and speed. If I were plugging in numbers for this, my first choices would be the simplest kindergarten possibilities --- x = 1 & y = 4, then x = 4 & y = 1, then x = 2 & y = 2 --- in the first 10-20 seconds, you should be able to eliminate at least three answer choices with very easy selections for the variable values. Focus speed & efficiency, not on making ideal choices. Once you are down to two answers, you may need a more sophisticated choice --- for example, none of those choices would eliminate (D), so you would have to choose something like x = 2 & y = 8. That eliminates everything except the OA, (B), and I didn't use anything other than single-digit choices. Often you can remain with single digit math and get everything you need.

Does all this make sense? Mike

Absolutely it does!

I am gonna follow this- To start with the easy numbers and the strategy would be to eliminate 4 options rather that just looking for the right answer. _________________

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
02 Jul 2013, 07:23

Since x * y is divisible by 4 Only the following two options are possible.

1. Both x and y should have a factor of 2 in them, which means both must be even numbers. 2. In case the above scenario is false, then one of them must have 4 as a factor.

Note : The reason for the above is that 4 can be written as 4 * 1 or 2 * 2. Here, x and y should contribute to this split of 4 in some way. If we observe the two conditions i have mentioned above, it is clear that the conditions follow from the above split of 4.

Re: If x and y are positive integer and xy is divisible by 4 [#permalink]
05 Sep 2014, 11:35

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