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NEW!!! Tough and tricky exponents and roots questions [#permalink]
12 Jan 2012, 02:50

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Exponents and roots problems are very common on the GMAT. So, it's extremely important to know how to manipulate them, how to factor out, take roots, multiply, divide, etc. Below are 11 problems to test your skills. Please post your thought process/solutions along with the answers.

I'll post OA's with detailed solutions tomorrow. Good luck.

1. If 357^x*117^y=a, where x and y are positive integers, what is the units digit of a? (1) 100<y^2<x^2<169 (2) x^2-y^2=23

2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} an integer? (1) y is an even perfect square and z is an odd perfect cube. (2) \sqrt{z} is not an integer.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
14 Jan 2012, 14:46

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7. If x is a positive integer is \sqrt{x} an integer? (1) \sqrt{7*x} is an integer (2) \sqrt{9*x} is not an integer

Must know for the GMAT: if x is a positive integer then \sqrt{x} is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether x is a perfect square.

(1) \sqrt{7*x} is an integer --> x can not be a perfect square because if it is, for example if x=n^2 for some positive integer n then \sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}. Sufficient.

(2) \sqrt{9*x} is not an integer --> \sqrt{9*x}=3*\sqrt{x}\neq{integer} --> \sqrt{x}\neq{integer}. Sufficient.

(2) 27^{x^2}=\frac{3}{3^{3y^2+1}} --> 3^{3x^2}=\frac{3}{3^{3y^2}*3} --> 3^{3x^2}*3^{3y^2}=1 --> 3^{3x^2+3y^2}=1 --> 3x^2+3y^2=0 (the power of 3 must be zero in order this equation to hold true) --> x^2+y^2=0 the sum of two non-negative values is zero --> both x and y must be zero --> x=y=0 --> x^2+y^3=0. Sufficient.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
14 Jan 2012, 14:52

9

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Expert's post

9. If x, y and z are non-zero numbers, what is the value of \frac{x^3+y^3+z^3}{xyz}? (1) xyz=-6 (2) x+y+z=0

(1) xyz=-6 --> infinitely many combinations of x, y and z are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.

(2) x+y+z=0 --> x=-(y+z) --> substitute this value of x into the expression in the stem --> \frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}, as x=-(y+z) then: \frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3. Sufficient.

Must know for the GMAT:(x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3 and (x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3.

Since both x and y are integers (and x+y>0) then x-y=2 and x+y=2 --> x=2 and y=0 --> (x+y)^{xy}=2^0=1=odd, so the answer to the question is No. Sufficient. (Note that x-y=1 and x+y=4 --> x=2.5 and y=1.5 is not a valid scenario (solution) as both unknowns must be integers)

(2) 2^x+3^y=\sqrt[(x+y)]{25} --> obviously \sqrt[(x+y)]{25} must be an integer (since 2^x+3^y=integer) and as x+y=integer then the only solution is \sqrt[(x+y)]{25}=\sqrt[2]{25}=5 --> x+y=2. So, 2^x+3^y=5 --> two scenarios are possible: A. x=2 and y=0 (notice that x+y=2 holds true) --> 2^x+3^y=2^2+3^0=5, and in this case: (x+y)^{xy}=2^0=1=odd; B. x=1 and y=1 (notice that x+y=2 holds true) --> 2^x+3^y=2^1+3^1=5, and in this case: (x+y)^{xy}=2^1=2=even.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
14 Jan 2012, 14:56

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11. What is the value of xy? (1) 3^x*5^y=75 (2) 3^{(x-1)(y-2)}=1

Notice that we are not told that the x and y are integers.

(1) 3^x*5^y=75 --> if x and y are integers then as 75=3^1*5^2 then x=1 and y=2 BUT if they are not, then for any value of x there will exist some non-integer y to satisfy given expression and vise-versa (for example if y=1 then 3^x*5^y=3^x*5=75 --> 3^x=25 --> x=some \ irrational \ #\approx{2.9}). Not sufficient.

(2) 5^{(x-1)(y-2)}=1 --> (x-1)(y-2)=0 --> either x=1 and y is ANY number (including 2) or y=2 and x is ANY number (including 1). Not sufficient.

(1)+(2) If from (2) x=1 then from (1) 3^x*5^y=3*5^y=75 --> y=2 and if from (2) y=2 then from (1) 3^x*5^y=3^x*25=75 --> x=1. Thus x=1 and y=2. Sufficient.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
14 Jan 2012, 18:16

Bunuel wrote:

2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer? (1) y is an even perfect square and z is an odd perfect cube. (2) \sqrt{z} is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> xyz=2^2*3^3*5^2.

(1) y is an even perfect square and z is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and z=3^3 =odd perfect square then x must be a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. But if z=1^3 =odd perfect square then x could be 3^3 which makes \sqrt{x} not an integer. Not sufficient.

(2) \sqrt{z} is not an integer. Clearly insufficient.

(1)+(2) As from (1) \sqrt{z}\neq{integer} then z\neq{1}, therefore it must be 3^3 (from 1) --> x is a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. Sufficient.

Answer: C.

In the highlighted part above, a couple of possibilities are missing. If y is an even perfect square, then we know y is divisible by 2^2. It may or may not be divisible by 3^2 and 5^2 (two choices each way, so four possibilities). So y could have any of four values:

2^2 (2^2)(3^2) (2^2)(5^2) (2^2)(3^2)(5^2)

Each of these values is possible when z = 1. While it doesn't affect the answer, it is also possible for x to be equal to 3 here, if y is equal to (2^2)(3^2)(5^2) and z is equal to 1, and x can also be (3)(5^2). I think you also meant to write 'odd perfect cube' instead of 'odd perfect square' (blue highlights). _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
15 Jan 2012, 01:39

Expert's post

IanStewart wrote:

Bunuel wrote:

2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer? (1) y is an even perfect square and z is an odd perfect cube. (2) \sqrt{z} is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> xyz=2^2*3^3*5^2.

(1) y is an even perfect square and z is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and z=3^3 =odd perfect square then x must be a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. But if z=1^3 =odd perfect square then x could be 3^3 which makes \sqrt{x} not an integer. Not sufficient.

(2) \sqrt{z} is not an integer. Clearly insufficient.

(1)+(2) As from (1) \sqrt{z}\neq{integer} then z\neq{1}, therefore it must be 3^3 (from 1) --> x is a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. Sufficient.

Answer: C.

In the highlighted part above, a couple of possibilities are missing. If y is an even perfect square, then we know y is divisible by 2^2. It may or may not be divisible by 3^2 and 5^2 (two choices each way, so four possibilities). So y could have any of four values:

2^2 (2^2)(3^2) (2^2)(5^2) (2^2)(3^2)(5^2)

Each of these values is possible when z = 1. While it doesn't affect the answer, it is also possible for x to be equal to 3 here, if y is equal to (2^2)(3^2)(5^2) and z is equal to 1, and x can also be (3)(5^2). I think you also meant to write 'odd perfect cube' instead of 'odd perfect square' (blue highlights).

Yes, I know Ian. For (1) I just discussed two possible scenarios to get an YES and NO answers to discard this statement, rather than listing all possibilities as you did. Thank you though for elaborating more on other cases. _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
17 Jan 2012, 10:44

Bunuel wrote:

7. If x is a positive integer is \sqrt{x} an integer? (1) \sqrt{7*y} is an integer (2) \sqrt{9*x} is not an integer

Must know for the GMAT: if x is a positive integer then \sqrt{x} is either a positive integer itself or an irrational number. (It can not be some reduced fraction eg 7/3 or 1/2)

Also note that the question basically asks whether x is a perfect square.

(1) \sqrt{7*x} is an integer --> x can not be a perfect square because if it is, for example if x=n^2 for some positive integer n then \sqrt{7x}=\sqrt{7n^2}=n\sqrt{7}\neq{integer}. Sufficient.

(2) \sqrt{9*x} is not an integer --> \sqrt{9*x}=3*\sqrt{x}\neq{integer} --> \sqrt{x}\neq{integer}. Sufficient.

Answer: D.

Originally I chose B because there was no mentioned of Y in the original answer choice until I read the rest of your explanation and realized it was probably just a typo.

By the way thank you for compiling such a nice collection of Math problems Bunuel!

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
17 Mar 2012, 05:33

Expert's post

devinawilliam83 wrote:

Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent

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.

So, remember: even roots have only non-negative value on the GMAT.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
06 Apr 2012, 23:40

Bunuel wrote:

9. If x, y and z are non-zero numbers, what is the value of \frac{x^3+y^3+z^3}{xyz}? (1) xyz=-6 (2) x+y+z=0

(1) xyz=-6 --> infinitely many combinations of x, y and z are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.

(2) x+y+z=0 --> x=-(y+z) --> substitute this value of x into the expression in the stem --> \frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}, as x=-(y+z) then: \frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3. Sufficient.

Must know for the GMAT:(x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3 and (x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3.

Answer: B.

Actually there is a direct formula for cubes of three entities-

x^3+y^3+z^3 - 3*x*y*z = (x+y+z) ( x^2+y^2+z^2 - x*y - y*z - z*x) so if one can figure out either of the quantity on RHS is 0 then you have the answer for the Q given... _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
13 Apr 2012, 18:14

Expert's post

Bunuel wrote:

2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer? (1) y is an even perfect square and z is an odd perfect cube. (2) \sqrt{z} is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> xyz=2^2*3^3*5^2.

(1) y is an even perfect square and z is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and z=3^3=odd \ perfect \ square then x must be a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. But if z=1^3=odd \ perfect \ cube then x could be 3^3 which makes \sqrt{x} not an integer. Not sufficient.

(2) \sqrt{z} is not an integer. Clearly insufficient.

(1)+(2) As from (1) \sqrt{z}\neq{integer} then z\neq{1}, therefore it must be 3^3 (from 1) --> x is a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. Sufficient.

Answer: C.

Bunuel can you please elaborate from where we have these values. I missed the link . x=1 and z=1^3

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
14 Apr 2012, 11:30

Expert's post

carcass wrote:

Bunuel wrote:

2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer? (1) y is an even perfect square and z is an odd perfect cube. (2) \sqrt{z} is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> xyz=2^2*3^3*5^2.

(1) y is an even perfect square and z is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and z=3^3=odd \ perfect \ square then x must be a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. But if z=1^3=odd \ perfect \ cube then x could be 3^3 which makes \sqrt{x} not an integer. Not sufficient.

(2) \sqrt{z} is not an integer. Clearly insufficient.

(1)+(2) As from (1) \sqrt{z}\neq{integer} then z\neq{1}, therefore it must be 3^3 (from 1) --> x is a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. Sufficient.

Answer: C.

Bunuel can you please elaborate from where we have these values. I missed the link . x=1 and z=1^3

Thanks

Are you talking about statement 1? The values of x, y, and z analyzed there are just possible values that satisfy statement 1.

Given that: xyz=2^2*3^3*5^2 and (1) says that y is an even perfect square and z is an odd perfect cube.

Now, if for example y=2^2*5^2=even \ perfect \ square and z=1^3=odd \ perfect \ cube then x=3^3. You can apply the similar logic to obtain other possible values of x, y, and z. _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
27 Apr 2012, 10:46

Bunuel wrote:

devinawilliam83 wrote:

Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent

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.

So, remember: even roots have only non-negative value on the GMAT.

Hope it's clear.

but won't this leave one in a dilemma? do i take this info with a pinch of salt, or is gmat only interested in the principal square root?

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
27 Apr 2012, 10:48

Expert's post

nrmlvrm wrote:

Bunuel wrote:

devinawilliam83 wrote:

Bunuel cant sqrt[x] be negative eg .. sqrt[4] = + or - 2.. in that case how is a suffucuent

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.

So, remember: even roots have only non-negative value on the GMAT.

Hope it's clear.

but won't this leave one in a dilemma? do i take this info with a pinch of salt, or is gmat only interested in the principal square root?

No ambiguity there whatsoever: even roots have only non-negative value on the GMAT Again: \sqrt{25}=5, NOT +5 or -5. _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
12 Jan 2013, 11:06

Bunuel wrote:

2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer? (1) y is an even perfect square and z is an odd perfect cube. (2) \sqrt{z} is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> xyz=2^2*3^3*5^2.

(1) y is an even perfect square and z is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and z=3^3=odd \ perfect \ square then x must be a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. But if z=1^3=odd \ perfect \ cube then x could be 3^3 which makes \sqrt{x} not an integer. Not sufficient.

(2) \sqrt{z} is not an integer. Clearly insufficient.

(1)+(2) As from (1) \sqrt{z}\neq{integer} then z\neq{1}, therefore it must be 3^3 (from 1) --> x is a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. Sufficient.

Answer: C.

Hi..

Why Not B?

If root of Z is not integer, then z must be 3^3 and this make root x an integer...... Please let me know where I am wrong...

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
13 Jan 2013, 02:39

Expert's post

sagarsingh wrote:

Bunuel wrote:

2. If x, y, and z are positive integers and xyz=2,700. Is \sqrt{x} and integer? (1) y is an even perfect square and z is an odd perfect cube. (2) \sqrt{z} is not an integer.

Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of some other integer. For example 27=3^3, is a perfect cube.

Make prime factorization of 2,700 --> xyz=2^2*3^3*5^2.

(1) y is an even perfect square and z is an odd perfect cube --> if y is either 2^2 or 2^2*5^2 and z=3^3=odd \ perfect \ square then x must be a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. But if z=1^3=odd \ perfect \ cube then x could be 3^3 which makes \sqrt{x} not an integer. Not sufficient.

(2) \sqrt{z} is not an integer. Clearly insufficient.

(1)+(2) As from (1) \sqrt{z}\neq{integer} then z\neq{1}, therefore it must be 3^3 (from 1) --> x is a perfect square which makes \sqrt{x} an integer: x=5^2 or x=1. Sufficient.

Answer: C.

Hi..

Why Not B?

If root of Z is not integer, then z must be 3^3 and this make root x an integer...... Please let me know where I am wrong...

thanks.

There are other cases possible. For example, z=3, y=2*3*5^2 and x=2.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
03 Jun 2013, 06:58

Bunuel wrote:

7. If x is a positive integer is \sqrt{x} an integer? (1) \sqrt{7*x} is an integer (2) \sqrt{9*x} is not an integer

Hi Bunuel, I would like you to point my mistake in evaluating equation 1

Lets Say: \sqrt{7*x} = K(an integer) Squaring both sides 7*x= K^2 x = (K^2)/7 Now, Since X is a positive integer, this means that K^2 has to be a multiple of 7. If K^2 = 7 then, x = 1 and \sqrt{x}= Integer. But,as per your explanation, x can't be an integer. Where am I going wrong? Please help

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
03 Jun 2013, 07:01

Expert's post

imhimanshu wrote:

Bunuel wrote:

7. If x is a positive integer is \sqrt{x} an integer? (1) \sqrt{7*x} is an integer (2) \sqrt{9*x} is not an integer

Hi Bunuel, I would like you to point my mistake in evaluating equation 1

Lets Say: \sqrt{7*x} = K(an integer) Squaring both sides 7*x= K^2 x = (K^2)/7 Now, Since X is a positive integer, this means that K^2 has to be a multiple of 7. If K^2 = 7 then, x = 1 and \sqrt{x}= Integer. But,as per your explanation, x can't be an integer. Where am I going wrong? Please help

Regards, imhimanshu

If k^2=7, then k won't be an integer as we defined.

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