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Math Expert V
Joined: 02 Sep 2009
Posts: 58135
If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal  [#permalink]

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11 00:00

Difficulty:   65% (hard)

Question Stats: 63% (02:11) correct 37% (02:23) wrong based on 228 sessions

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

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

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Retired Moderator V
Joined: 27 Oct 2017
Posts: 1242
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: If a and b are integer and  [#permalink]

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2
$$500 = 5^3* 2^2$$
Also$$( a^(1/3) *b^(1/2))^6 = a^2*b^3$$
Hence a can be 2, b can be 5.
But a+b =7.
It is not in the option. Bingo: we have forgot the fact that even power changes negative to positive.
Actually a can be +/- 2 Now considering a = -2, b =5
Hence a+b = -5+2 =3. Answer B
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Current Student P
Joined: 18 Aug 2016
Posts: 613
Concentration: Strategy, Technology
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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1
1
Bunuel wrote:
If a and b are integers and $$(\sqrt{a}*\sqrt{b})^6 = 500$$, 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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Senior SC Moderator V
Joined: 22 May 2016
Posts: 3451
If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal  [#permalink]

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1
Bunuel wrote:
If a and b are integers and $$(\sqrt{a}*\sqrt{b})^6 = 500$$, then a + b could equal

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

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

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
$$a = -2$$

$$b^3 = 5^3$$

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

$$b = 5$$

$$a + b$$?

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

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Joined: 15 Oct 2017
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Location: India
Concentration: Entrepreneurship, Marketing
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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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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)
Director  P
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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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Bunuel wrote:
If a and b are integers and $$(\sqrt{a}*\sqrt{b})^6 = 500$$, 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

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

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Bunuel wrote:
If a and b are integers and $$(\sqrt{a}*\sqrt{b})^6 = 500$$, 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: $$(\sqrt{a}*\sqrt{b})^6 = 500$$ (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:
$$(\sqrt{a}*\sqrt{b})^6 = (a^\frac{1}{3} * b^\frac{1}{2})^6 = a^2 * b^3 = 500$$ (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.
Target Test Prep Representative G
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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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Bunuel wrote:
If a and b are integers and $$(\sqrt{a}*\sqrt{b})^6 = 500$$, 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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Director  P
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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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Bunuel wrote:
If a and b are integers and $$(\sqrt{a}*\sqrt{b})^6 = 500$$, then a + b could equal

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

The given equation can be written as,

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

So, $$a + b = 7$$ or $$3.$$ Hence, option B.

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Joined: 31 Dec 2017
Posts: 28
Concentration: Finance
Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal  [#permalink]

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Bunuel wrote:
If a and b are integers and $$(\sqrt{a}*\sqrt{b})^6 = 500$$, then a + b could equal

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

$$(\sqrt{a}*\sqrt{b})^6 = [\sqrt{a}]^6 * [\sqrt{b}]^6 = [a]^{\frac{6}{3}} * [b]^{\frac{6}{2}} = a^2 * b^3 = 500.$$

$$500 = (2)^2 * (5)^3 = (-2)^2 * (5)^3$$

$$(a+b)$$ could equal $$(2+5) = 7$$, or $$(-2+5) = 3$$. Ans - B.
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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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