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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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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, 01:59
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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

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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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21 Nov 2017, 02:41
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)
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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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21 Nov 2017, 02:43
1
1
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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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, 03:36
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

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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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21 Nov 2017, 19:13
1
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
$$-2 + 5 = 3$$

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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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22 Nov 2017, 05:49
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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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, 11:49
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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Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal  [#permalink]

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27 Jan 2018, 08:28
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

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

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27 Jan 2018, 19:32
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

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

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

$$(a+b)$$ could equal $$(2+5) = 7$$, or $$(-2+5) = 3$$. Ans - B.
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Re: If a and b are integer and  [#permalink]

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20 Apr 2018, 06:14
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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Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal  [#permalink]

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21 Apr 2019, 00:50
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Re: If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal   [#permalink] 21 Apr 2019, 00:50
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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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