Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

10. If \(x\) and \(y\) are non-negative integers and \(x+y>0\) is \((x+y)^{xy}\) an even integer? (1) \(2^{x-y}=\sqrt[(x+y)]{16}\) (2) \(2^x+3^y=\sqrt[(x+y)]{25}\)

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.

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 \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).

10. If \(x\) and \(y\) are non-negative integers and \(x+y>0\) is \((x+y)^{xy}\) an even integer? (1) \(2^{x-y}=\sqrt[(x+y)]{16}\) (2) \(2^x+3^y=\sqrt[(x+y)]{25}\)

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: \((x+y)^{xy}=2^1=2=even\).

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.

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). _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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]

Show Tags

17 Jan 2012, 11:44

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!

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]

Show Tags

07 Apr 2012, 00:40

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]

Show Tags

13 Apr 2012, 19:14

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

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]

Show Tags

27 Apr 2012, 11:46

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?

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]

Show Tags

12 Jan 2013, 12:06

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...

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]

Show Tags

03 Jun 2013, 07:58

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

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.

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...