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# Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d

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Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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Updated on: 22 May 2018, 05:22
00:00

Difficulty:

35% (medium)

Question Stats:

69% (00:57) correct 31% (01:05) wrong based on 85 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

I dont understand the answer. If you ask me it's (E) --> "Statements (1) and (2) TOGETHER are NOT sufficient"

The question does not insist on "integers" so I was thinking more in terms of negative roots. here's the explanation:
If a= [square_root]5, b=[square_root](-5), c=[square_root](7), d=[square_root]-(7) , it still satisfies Eqn#1 and the product of abcd is not even(abcd= 1225).

When do you decide you can assume the problem involves integers and does not involve integers?

M03-35

Originally posted by chethanjs on 05 Apr 2011, 18:44.
Last edited by pikolo2510 on 22 May 2018, 05:22, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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05 Apr 2011, 20:38
is the first equation equal zero? Something is missing in the first equation i guess.
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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05 Apr 2011, 21:27
Yes, the first option has a^2+ b^2 + c^2 + d^2 = 0, which automatically makes A as correct answer. GMAT does not deal with complex numbers, and only real numbers are in consideration here. so square of any real number is non-negative and hence a = b = c = d = 0.

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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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06 Apr 2011, 03:04
Sorry, yea, 1st eqn is a^2+ b^2 + c^2 + d^2 = 0 .

@subash: oh ok. Thanks.
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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07 Apr 2011, 03:33
Wouldn't statement 2 be sufficient as well?

If a = b = c = d, then abcd = a^4.

If a is positive then, a^4 is positive.

If a is negative, a^4 is still positive.

Only remaining case is a = 0, but I assume we can think of 0 as positive.
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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07 Apr 2011, 03:38
1
oster wrote:
Wouldn't statement 2 be sufficient as well?

If a = b = c = d, then abcd = a^4.

If a is positive then, a^4 is positive.

If a is negative, a^4 is still positive.

Only remaining case is a = 0, but I assume we can think of 0 as positive.

Question asks whether a.b.c.d is EVEN.

a=1,
a.b.c.d=a^4=1; odd

a=2,
a.b.c.d=a^4=16; even

a=0,
a.b.c.d=a^4=0; even

By the way; "0" is neither +ve nor -ve.
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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07 Apr 2011, 11:04
fluke wrote:
oster wrote:
Wouldn't statement 2 be sufficient as well?

If a = b = c = d, then abcd = a^4.

If a is positive then, a^4 is positive.

If a is negative, a^4 is still positive.

Only remaining case is a = 0, but I assume we can think of 0 as positive.

Question asks whether a.b.c.d is EVEN.

a=1,
a.b.c.d=a^4=1; odd

a=2,
a.b.c.d=a^4=16; even

a=0,
a.b.c.d=a^4=0; even

By the way; "0" is neither +ve nor -ve.

Ah yes, d'oh!

Thanks!
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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07 Apr 2011, 13:20
I am still confused. Wouldn't it be correct to presume that the statement 1)a^2+ b^2 + c^2 + d^2 = 0 would be true only if a, b, c, d equal zero??

As follows, abcd would equal zero and thus answer A) would be correct
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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07 Apr 2011, 13:28
katealpha wrote:
I am still confused. Wouldn't it be correct to presume that the statement 1)a^2+ b^2 + c^2 + d^2 = 0 would be true only if a, b, c, d equal zero??

As follows, abcd would equal zero and thus answer A) would be correct

katealpha, you scored a bull's eye. No need to be confused.
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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08 Apr 2011, 11:47
katealpha wrote:
I am still confused. Wouldn't it be correct to presume that the statement 1)a^2+ b^2 + c^2 + d^2 = 0 would be true only if a, b, c, d equal zero??

As follows, abcd would equal zero and thus answer A) would be correct

You're correct. any square is always $$>= 0$$. The only case when a bunch of squares will add to zero is when all squares are themselves zero and this will only happen when each number is zero.
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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d  [#permalink]

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22 May 2018, 04:48
chethanjs wrote:
Is the product abcd even?

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

I dont understand the answer. If you ask me it's (E) --> "Statements (1) and (2) TOGETHER are NOT sufficient"

The question does not insist on "integers" so I was thinking more in terms of negative roots. here's the explanation:
If a= [square_root]5, b=[square_root](-5), c=[square_root](7), d=[square_root]-(7) , it still satisfies Eqn#1 and the product of abcd is not even(abcd= 1225).

When do you decide you can assume the problem involves integers and does not involve integers?

M03-35

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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Re: Is the product abcd even? (1) a^2+ b^2 + c^2 + d^2 (2) a = b = c = d &nbs [#permalink] 22 May 2018, 04:48
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