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# If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal

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Math Expert
Joined: 02 Sep 2009
Posts: 42605

Kudos [?]: 135629 [0], given: 12705

If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink]

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21 Nov 2017, 00:59
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Difficulty:

65% (hard)

Question Stats:

55% (01:39) correct 45% (01:11) wrong based on 72 sessions

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If a and b are integers and $$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$, then a + b could equal

A. 2
B. 3
C. 4
D. 5
E. 6
[Reveal] Spoiler: OA

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Kudos [?]: 135629 [0], given: 12705

Intern
Joined: 15 Oct 2017
Posts: 7

Kudos [?]: 5 [0], given: 50

Location: India
Concentration: Entrepreneurship, Marketing
Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink]

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21 Nov 2017, 01:41
By solving LHS, we get a^2*b^3=500
We know 500 = 2^2*5^3

Hence, a=2 b=5
a+b = 7 (Not in the options)

Kudos [?]: 5 [0], given: 50

Director
Joined: 18 Aug 2016
Posts: 598

Kudos [?]: 181 [0], given: 138

GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink]

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21 Nov 2017, 01:43
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Bunuel wrote:
If a and b are integers and $$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$, then a + b could equal

A. 2
B. 3
C. 4
D. 5
E. 6

a^2 * b^3

now a =2 or -2 and b = 5 (125*4)
-2+5 = 3

B
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Luckisnoexcuse

Kudos [?]: 181 [0], given: 138

Manager
Joined: 31 Jul 2017
Posts: 118

Kudos [?]: 17 [0], given: 325

Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink]

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21 Nov 2017, 02:36
Bunuel wrote:
If a and b are integers and $$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$, then a + b could equal

A. 2
B. 3
C. 4
D. 5
E. 6

a^2 * b^3 = (-2)^2 * 5^3 = 2^2 * 5^3

a + b = 3/7

Sent from my Lenovo P1a42 using GMAT Club Forum mobile app

Kudos [?]: 17 [0], given: 325

VP
Joined: 22 May 2016
Posts: 1126

Kudos [?]: 402 [0], given: 645

If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink]

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21 Nov 2017, 18:13
Bunuel wrote:
If a and b are integers and $$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$, then a + b could equal

A. 2
B. 3
C. 4
D. 5
E. 6

$$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$

Do the prime factorization for 500: $$2^25^3$$

Rewrite
$$(a^{\frac{1}{3}} * b^{\frac{1}{2}})^6 = 2^2*5^3$$

Distribute the exponent

$$a^{(\frac{1}{3}*6)} * b^{(\frac{1}{2}*6)} = 2^2*5^3$$

$$a^2 * b^3 = 2^2 * 5^3$$

$$a^2 = 2^2 = 4$$

$$\sqrt{a^2} =\sqrt{4}$$
$$a = 2$$ OR
$$a = -2$$

$$b^3 = 5^3$$

$$(\sqrt[3]{b^3}) = (\sqrt[3]{5^3}$$)

$$b = 5$$

$$a + b$$?

$$2 + 5 = 7$$ Not an answer choice
$$-2 + 5 = 3$$

Kudos [?]: 402 [0], given: 645

Intern
Status: Prepping for GMAT
Joined: 06 Nov 2017
Posts: 16

Kudos [?]: 5 [0], given: 26

Location: France
GPA: 4
If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink]

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22 Nov 2017, 04:49
Bunuel wrote:
If a and b are integers and $$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$, then a + b could equal

A. 2
B. 3
C. 4
D. 5
E. 6

Before people complain that the answer they have found is not in the suggestions of the problem prompt, they need to remember that an integer can be either positive and negative. This was, in my opinion, the main trap with this problem.

We have: $$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$ (A)

This equality is too complex for my tastes, let's simplify it while keeping in mind that:

1/ $$\sqrt{x} = x^\frac{1}{2}$$
2/ $$\sqrt[y]{x} = x^\frac{1}{y}$$
3/ $$(a*b)^c = a^c * b^c$$

Therefore, using the first 3 properties above, A becomes:
$$(\sqrt[3]{a}*\sqrt{b})^6 = (a^\frac{1}{3} * b^\frac{1}{2})^6 = a^2 * b^3 = 500$$ (B)

If we decompose 500 into its prime components, we get $$500 = 2^2*5^3$$

Thus B can be written as: $$a^2 * b^3 = 2^2*5^3$$

At this stage, we're tempted to say that $$a = 2$$ and $$b = 5$$ and thus complain that $$a+b = 7$$ isn't in the choices.

Or, we can remember that even exponents remove the negative sign from negative integers and deduce that the only solution possible is $$a = -2$$ and $$b = 5$$ (since odd exponents keep the negative sign of negative integers and since 500 is positive, it follows that b can never be negative).

Thus, $$a + b = 3$$ i.e. answer B.

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Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
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Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal [#permalink]

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27 Nov 2017, 10:49
Bunuel wrote:
If a and b are integers and $$(\sqrt[3]{a}*\sqrt{b})^6 = 500$$, then a + b could equal

A. 2
B. 3
C. 4
D. 5
E. 6

We can rewrite the expression:

[(a^1/3) x (b^1/2)]^6 = 500

a^2 x b^3 = 500

a^2 x b^3 = 2^2 x 5^3

Thus a could be 2 and b could be 5, and a + b = 7. However, notice that a^2 = 2^2 = 4, so a could be -2 also. In that case, a + b = -2 + 5 = 3.

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Jeffery Miller

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Kudos [?]: 981 [0], given: 5

Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal   [#permalink] 27 Nov 2017, 10:49
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