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# If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the

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Manager
Joined: 27 Feb 2010
Posts: 88
Location: Denver
If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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26 Apr 2010, 21:07
10
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Difficulty:

35% (medium)

Question Stats:

65% (00:40) correct 35% (00:47) wrong based on 1939 sessions

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If $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$, then the value of M is:

A. Less than 3
B. Equal to 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4
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Posts: 52294
Re: The value of M ?  [#permalink]

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27 Apr 2010, 07:06
105
1
85
zz0vlb wrote:
If $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$, then the value of M is:

A. less than 3
B. equal to 3
C. between 3 and 4
D. equal to 4
E. greater than 4

Here is a little trick: any positive integer root from a number more than 1 will be more than 1.

For instance: $$\sqrt[1000]{2}>1$$.

Hence $$\sqrt[3]{4}>1$$ and $$\sqrt[4]{4}>1$$ --> $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}=2+(number \ more \ then \ 1)+(number \ more \ then \ 1)>4$$

Answer: E.
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Manager
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Posts: 233
Re: The value of M ?  [#permalink]

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26 Apr 2010, 21:52
8
3
zz0vlb wrote:
Find the value of M (see attachment). I want to see other approaches to this problem.

Source: GMAT Prep

sqrt(4) = 2
sqrt(sqrt(4)) = 1.414 approx
hence sqrt(4) + sqrt(sqrt(4)) = 3.414
cuberoot(4) > 1 atleast
hence answer is M>4.
##### General Discussion
Manager
Joined: 27 Feb 2010
Posts: 88
Location: Denver
Re: The value of M ?  [#permalink]

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27 Apr 2010, 08:06
1
THANKS Bunuel, this is exactly what i need to know to attack such similar questions..
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Re: Value of M?  [#permalink]

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23 Sep 2010, 11:09
7
2
udaymathapati wrote:
Please explain the answer.

Attachment:
Image2.JPG

$$M=4^{1/2} + 4^{1/3} + 4^{1/4}$$

Now we know that $$4^{1/2} = 2$$

We also know that $$4^{1/4} = \sqrt{2} \approx 1.414 > 1$$

And finally $$4^{1/3} > 4^{1/4} \Rightarrow 4^{1/3}>1$$

So combining all three together $$M > 2+1+1 \Rightarrow M > 4$$
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Re: The value of M ?  [#permalink]

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23 Sep 2010, 19:16
1
Bunuel wrote:
zz0vlb wrote:
If $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$, then the value of M is:

A. less than 3
B. equal to 3
C. between 3 and 4
D. equal to 4
E. greater than 4

Here is a little trick: any positive integer root from a number more than 1 will be more than 1.

For instance: $$\sqrt[1000]{2}>1$$.

Hence $$\sqrt[3]{4}>1$$ and $$\sqrt[4]{4}>1$$ --> $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}=2+(number \ more \ then \ 1)+(number \ more \ then \ 1)>4$$

Answer: E.

got the correct answer, however thanks Bunuel for trick....
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Manager
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Re: GMAT PREP STATS question  [#permalink]

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08 Apr 2012, 01:57
piyushksharma wrote:
can someone explain how to do this type of questions in less time.
If M = (4)^.5 + (4)^.3 + (4)^.25
, then the value of M is
1)less than 3
2) equal to 3
3)between 3 and 4
4)equal to 4
5)greater than 4

here \sqrt{4}=2
1<4^(1/3)<2
1<4^(1/4)<2

hence the sum, S;

4<S

P.S.: GO FOR THE FIXED MINIMUM VALUE THE ANSWER WILL BECOME EASY.

Hope this helps...!!
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Joined: 23 Feb 2012
Posts: 8
Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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21 Apr 2012, 12:14
1
I was wondering why we are not considering the negative roots of 4 in this case. For instance, sq root of 4 would be 2 and -2...same for the 4th root... Any rule / trick that I might be missing here?

Look forward to the answer.

Cheers!
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Posts: 52294
Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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21 Apr 2012, 12:18
4
4
immune wrote:
I was wondering why we are not considering the negative roots of 4 in this case. For instance, sq root of 4 would be 2 and -2...same for the 4th root... Any rule / trick that I might be missing here?

Look forward to the answer.

Cheers!

Welcome to GMAT Club. Below might help to clear your doubts.

1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a unique non-negative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, $$\sqrt{25}=+5$$ and $$-\sqrt{25}=-5$$. Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html
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⁴√(4)+√(4)+³√(4)  [#permalink]

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10 Jan 2013, 01:33
All,

I noted down this question, but I forgot the source and cannot retrieve the correct answer. Needless to say the question asked for the approximate value of the equation & whether it was between, or greater/smaller than a combination of numbers, i.e. +/- 3 & 3/4, 3 & 4 and so forth. The question is whether there is a shortcut for simplifying this or looking at it under a different perspective other than sheer number sense, Thanks and apologies for being vague.

⁴√(4)+√(4)+³√(4)
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Posts: 371
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Re: ⁴√(4)+√(4)+³√(4)  [#permalink]

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10 Jan 2013, 02:01
As the problem expects approximate value between nearest integers, simplify as below:

⁴√(4)+√(4)+³√(4) = √(2) + 2 +³√(4) = 1.41 + 2 + 1.44 = 4.85

The answer should be between 4 & 5.
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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28 Jan 2013, 07:46
Thanks for this wonderful trick bunuel!
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If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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21 Jun 2013, 07:47
Similar question to practice: http://gmatclub.com/forum/new-tough-and ... l#p1029227
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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10 Oct 2013, 21:14
Did in this way:

4 ^1/2 + 4 ^ 1/3 + 4 ^ 1/4

= 4 ^1/2 ( 1 + 4 ^2/3 + 4 ^1/2)

= 2 (1 + 2 + 4 ^2/3)

The above value is obviously above 6, so answer = E
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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11 Oct 2013, 04:13
2
somalwar wrote:
Did in this way:

4 ^1/2 + 4 ^ 1/3 + 4 ^ 1/4

= 4 ^1/2 ( 1 + 4 ^2/3 + 4 ^1/2)

= 2 (1 + 2 + 4 ^2/3)

The above value is obviously above 6, so answer = E

Factoring out is not correct:
4^(1/2)*4^(2/3) does not equal to 4^(1/3).
4^(1/2)*4^(1/2) does not equal to 4^(1/4).

$$a^n*a^m=a^{n+m}$$ not $$a^{nm}$$.

Hope it helps.
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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12 Oct 2013, 23:35
Thank you Bunuel for pointing it out..... my mistake
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If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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02 Nov 2014, 20:05
Bunuel wrote:
somalwar wrote:
Did in this way:

4 ^1/2 + 4 ^ 1/3 + 4 ^ 1/4

= 4 ^1/2 ( 1 + 4 ^2/3 + 4 ^1/2)

= 2 (1 + 2 + 4 ^2/3)

The above value is obviously above 6, so answer = E

Factoring out is not correct:
4^(1/2)*4^(2/3) does not equal to 4^(1/3).
4^(1/2)*4^(1/2) does not equal to 4^(1/4).

$$a^n*a^m=a^{n+m}$$ not $$a^{nm}$$.

Hope it helps.

Is there any way to factor out this problem?

I'm guessing not ... I've been trying to for a while now and I can't.

4^(1/2) = 2^1
4^(1/3) = 2^(2/3)
4^(1/4) = 2^(2/4)

Therefore 2( 1^1 + 1^(2/3) + 1^(2/4) )

HOWEVER since 1^(ANYTHING) = 1 this solution will not work.
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Joined: 10 Dec 2014
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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17 Dec 2014, 04:20
ak1802 wrote:
Bunuel wrote:
somalwar wrote:
Did in this way:

4 ^1/2 + 4 ^ 1/3 + 4 ^ 1/4

= 4 ^1/2 ( 1 + 4 ^2/3 + 4 ^1/2)

= 2 (1 + 2 + 4 ^2/3)

The above value is obviously above 6, so answer = E

Factoring out is not correct:
4^(1/2)*4^(2/3) does not equal to 4^(1/3).
4^(1/2)*4^(1/2) does not equal to 4^(1/4).

$$a^n*a^m=a^{n+m}$$ not $$a^{nm}$$.

Hope it helps.

Is there any way to factor out this problem?

I'm guessing not ... I've been trying to for a while now and I can't.

4^(1/2) = 2^1
4^(1/3) = 2^(2/3)
4^(1/4) = 2^(2/4)

Therefore 2( 1^1 + 1^(2/3) + 1^(2/4) )

HOWEVER since 1^(ANYTHING) = 1 this solution will not work.

Is taking 2 common the right way to simplify??
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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17 Dec 2014, 18:06
LUC999 wrote:
ak1802 wrote:
Bunuel wrote:
Factoring out is not correct:
4^(1/2)*4^(2/3) does not equal to 4^(1/3).
4^(1/2)*4^(1/2) does not equal to 4^(1/4).

$$a^n*a^m=a^{n+m}$$ not $$a^{nm}$$.

Hope it helps.

Is there any way to factor out this problem?

I'm guessing not ... I've been trying to for a while now and I can't.

4^(1/2) = 2^1
4^(1/3) = 2^(2/3)
4^(1/4) = 2^(2/4)

Therefore 2( 1^1 + 1^(2/3) + 1^(2/4) )

HOWEVER since 1^(ANYTHING) = 1 this solution will not work.

Is taking 2 common the right way to simplify??

That's not the correct way.

Bunuel's method would be the best to solve it.
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If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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19 Dec 2014, 03:28
PareshGmat wrote:

That's not the correct way.

you can solve this question, by factoring the 2 out of the given expression. check out the following explanation

$$M=$$$$4 ^{1/2} + 4 ^{1/3} + 4 ^{1/4}$$

=$$2 + 2^{2/3} + 2^{2/4}$$

=$$2(1+ 2^{2/3-1} + 2^{1/2-1})$$

= $$2(1+ 2^{-1/3} + 2^{-1/2})$$

$$2^{-1/3}$$$$\approx 0.8$$

$$2^{-1/2}$$$$\approx 0.7$$

$$=2(1+0.8 +0.7)$$

$$=2(2.5)$$

$$=5$$
If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the &nbs [#permalink] 19 Dec 2014, 03:28

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# If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the

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