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# M03-35

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:21
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5% (low)

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76% (00:44) correct 24% (01:07) wrong based on 212 sessions

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Is the product $$abcd$$ even?

(1) $$a^2 + b^2 + c^2 + d^2 = 0$$

(2) $$a = b = c = d$$

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16 Sep 2014, 00:21
1
Official Solution:

Statement (1) by itself is sufficient. All the variables equal zero, and the product of the variables is zero; therefore their product is even.

Statement (2) by itself is insufficient. The variables can be either odd or even. If all the variables are even, their product is even; if they are odd, their product is odd.

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17 Jun 2015, 21:57
The key here is to realize that zero is even!
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20 Sep 2015, 04:22
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. All the variables equal zero, and the product of the variables is zero; therefore their product is even.

Statement (2) by itself is insufficient. The variables can be either odd or even. If all the variables are even, their product is even; if they are odd, their product is odd.

Hi Bunuel, Please elaborate on this:
"All the variables equal zero, and the product of the variables is zero;"
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20 Sep 2015, 04:40
1
scofield1521 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. All the variables equal zero, and the product of the variables is zero; therefore their product is even.

Statement (2) by itself is insufficient. The variables can be either odd or even. If all the variables are even, their product is even; if they are odd, their product is odd.

Hi Bunuel, Please elaborate on this:
"All the variables equal zero, and the product of the variables is zero;"

Hi,
if i may help
all four terms in $$a^2+b^2+c^2+d^2$$ are positive as square of any number is always positive, so only possibility for the sum to be zero is when all four terms are 0, otherwise the sum will be some positive integer or positive fraction...
zero is even so suff...
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14 Apr 2016, 12:10
We should not consider irrational numbers?
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14 Apr 2016, 12:32
forsellingonline1 wrote:
We should not consider irrational numbers?

The square of an irrational number is still positive, so the sum of the squares of 4 irrational numbers will be positive, not 0.
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26 Jun 2017, 07:09
I'm getting lost in all this, can't a and b for example =-1 and c and d equal 1? Wouldn't the sum still be 0? Because 1^2=1 but -1^2=-1. I don't understand when we have to assume that all the numbers must be positive or can be either.
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26 Jun 2017, 08:08
stan3544 wrote:
I'm getting lost in all this, can't a and b for example =-1 and c and d equal 1? Wouldn't the sum still be 0? Because 1^2=1 but -1^2=-1. I don't understand when we have to assume that all the numbers must be positive or can be either.

The square of a number is always more than or equal to 0.

So, if b = -1, then b^2 = (-1)^2 = 1.

Is the product abcd even?

(1) a^2+b^2+c^2+d^2=0 --> number squared is always non-negative (zero or positive), so the sum of 4 non-negative values to be 0 then each must be zero, so abcd=0=even. Sufficient.

(2) a=b=c=d. Clearly insufficient.

Hope it's clear.
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04 Nov 2018, 20:30
Hi Bunuel,

What if both a and b are square root of negative one (-1) and both c and d are square root of positive one (1),
then the answer of first statement will still be zero.
Are we supposed to consider root of negative numbers?
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04 Nov 2018, 21:36
HrusheekeshJoshi wrote:
Hi Bunuel,

What if both a and b are square root of negative one (-1) and both c and d are square root of positive one (1),
then the answer of first statement will still be zero.
Are we supposed to consider root of negative numbers?

Even roots from negative numbers are not defined on the GMAT. All numbers on the GMAT are real numbers.
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11 Jul 2019, 07:35
I will keep this one relatively short, writing in a different style from my typical responses. The issue here is that we need to assess whether a certain product, abcd, is even. I am surprised that the problem does not state that we are working with integers, but I guess that is presupposed by the question itself (since decimals or fractions cannot be considered even or odd). The fact that there are four numbers instead of, say, two is irrelevant. All we need to do is consider the basic product relationships between even and odd numbers:

odd x odd = odd (e.g., 1 x 3 = 3)
even x even = even (e.g., 2 x 2 = 4)
odd x even = even (e.g., 1 x 2 = 2)

If we understand that a product involving four numbers is the same as two sets of a product of two numbers, ([#][#]) x ([#][#]), then the groups would follow the same product relationship outlined above. We just need to keep tabs on how many of the four unknowns are odd or even.

In terms of the two statements, the second one would likely appear simpler than the first to most people, if for no other reason than that statement (1) requires more symbols to interpret. If a = b = c = d, then we can either think in terms of odds or evens, or we can substitute numbers themselves. Keep it simple: an odd is an odd for the purposes of the relationship in the question, so there is no need to choose different numbers.

odd x odd x odd x odd = odd (e.g, 1 x 1 x 1 x 1 = 1)
even x even x even x even = even (e.g., 2 x 2 x 2 x 2 = 16)

Clearly, statement (2) is NOT sufficient, yielding one type of answer or another depending on whether the unknowns are odd or even, so we can scrap choices (B) and (D). What about statement (1)? If the sum of the squares of four numbers is 0, then all the numbers have to be 0 itself. Why? First, the complex number i does not appear on the GMAT™--you have to understand the rules of the test you are taking. Second, regardless of whether any number is positive or negative, given the previous statement about i, as soon as you square it, it cannot be negative anymore. Before you might say that the calculator tells you that, for instance, -2^2 = -4, consider that the calculator is following the order of operations, taking 2, squaring it, and then applying the negative sign. If you type in the number correctly, as in (-2), and square it, you will most definitely get positive 4. Now it should be clear why all the numbers in this problem must be 0, which happens to be an even number. (There is no separate category for 0, in terms of evens or odds.) Statement (1) is sufficient, so (A) is the answer.
Re: M03-35   [#permalink] 11 Jul 2019, 07:35
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# M03-35

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