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NEW!!! Tough and tricky exponents and roots questions [#permalink]

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12 Jan 2012, 03:50

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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.

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) \(2^x+3^y=\sqrt[(x+y)]{25}\)

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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04 Aug 2013, 15:30

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 Bunuel,

Is it not true to say that 2^2*5^2 could be called a perfect square too? In that case, from statement 1 and 2 one possibility is that x=2^2*5^2 apart from x=1 and x=5^2.

I know in this case it wont make a difference to the answer but I want to clarify the concept.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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05 Sep 2013, 12:30

Bunuel wrote:

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) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.

Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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05 Sep 2013, 12:35

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Expert's post

Acropora wrote:

Bunuel wrote:

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) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.

Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?

23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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05 Sep 2013, 13:45

Bunuel wrote:

Acropora wrote:

Bunuel wrote:

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) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.

Can anybody clarify for me how X-Y = 1 in the 2nd part of the question?

23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

Does this make sense?

Ooo I got it. Man.. I can't believe I didn't realize that. Thanks!

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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07 Sep 2013, 00:38

we are asked only about units digit so the problem simply can be seen as (7^x)*(7^y) that is nothing but 7^(x+y) from option a) x,y are 11,12 from option b)(x+y)*(x-y)=23(prime number) thus a+b=23 thus the units digit can be inferred from both the options

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Oct 2013, 00:26

IanStewart wrote:

Bunuel wrote:

3. If \(x>y>0\) then what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)? (1) \(x+y=4+2\sqrt{xy}\) (2) \(x-y=9\)

Since in both of the posts above, the answer to this question was given incorrectly, I thought I'd post a quick solution. Using the difference of squares,

So if we can find the value of \(\sqrt{x} - \sqrt{y}\), we can answer the question.

Now from Statement 1, we have

\(\begin{align} x + y &= 4 + 2\sqrt{xy} \\ x - 2\sqrt{xy} + y &= 4 \\ (\sqrt{x} - \sqrt{y})^2 &= 4 \\ \sqrt{x} - \sqrt{y} &= 2 \end{align}\) (here we know the root is 2, and not -2, since x > y). So Statement 1 is sufficient.

in the question could not demistify the denominator ,otherwise was on track. thanks for the help.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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07 Oct 2013, 01:11

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.

Dear Mr. Bunuel, As you say, a perfect cube is an integer that can be written as the cube of some OTHER integer. I am wondering whether 1^3 can not be a perfect cube in accordance with this definition. Thanks for your support

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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07 Oct 2013, 01:15

Expert's post

smallurgo 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.

Dear Mr. Bunuel, As you say, a perfect cube is an integer that can be written as the cube of some OTHER integer. I am wondering whether 1^3 can not be a perfect cube in accordance with this definition. Thanks for your support

Yes, 1 is a perfect cube and perfect square. _________________

because I got x^6+y^9=0 --> (x^2)^3 + (y^3)^3=0 --> (x^2)^3 = - (y^3)^3. The difference is in "-" sing in front of (y^3)^3. So if there is a "-" sign in front of it, does it give us the same result?

Thanks _________________

Please don't forget to give Kudos if you like my post! :-)

because I got x^6+y^9=0 --> (x^2)^3 + (y^3)^3=0 --> (x^2)^3 = - (y^3)^3. The difference is in "-" sing in front of (y^3)^3. So if there is a "-" sign in front of it, does it give us the same result?

Thanks

Both give the same result: \(x^2=-y^3\). _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 Apr 2014, 10:08

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 Bunuel Can you please explain second statement. How did you figured out "\(\sqrt{z}\) is not an integer"

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 Apr 2014, 11:36

Expert's post

282552 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 Bunuel Can you please explain second statement. How did you figured out "\(\sqrt{z}\) is not an integer"

I did not deduce that myself, it's given in the second statement: "(2) \(\sqrt{z}\) is not an integer." _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 Apr 2014, 11:54

Bunuel wrote:

282552 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 Bunuel Can you please explain second statement. How did you figured out "\(\sqrt{z}\) is not an integer"

I did not deduce that myself, it's given in the second statement: "(2) \(\sqrt{z}\) is not an integer."

I am sorry I wanted to ask According to second statement why can't z be let's say 3 or 3^3 or anything else. In that case we would be able to tell that we can have a perfect square for x

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 Apr 2014, 12:34

Expert's post

Bunuel wrote:

282552 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. I did not deduce that myself, it's given in the second statement: "(2) \(\sqrt{z}\) is not an integer."

I am sorry I wanted to ask According to second statement why can't z be let's say 3 or 3^3 or anything else. In that case we would be able to tell that we can have a perfect square for x

Are you talking about the first statement??? Or when we consider the statements together???

(1) says that \(z\) is an odd perfect cube, thus it can be only 1 or 3^3. _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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28 Apr 2014, 13:10

Hi,

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

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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29 Apr 2014, 07:46

Expert's post

sheolokesh wrote:

Hi,

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

when you explained the Stmt 1, I found it confusing.. However I took it in a different way. Let me know if my thinking is right.. For the 1st stmt to be right, x must be 7. So sqrt(7) is not a integer.. Sufficient..

I really missed the 2nd stmt when I did it.. Your way of solving and explaining is really good.. keep us posted with a more and more like these..

For (1) x=7 is not the only possible value to satisfy \(\sqrt{7*x}=integer\). For example, x can also be 7*4, 7*9, 7*16, ... basically any number of a form 7*integer^2. _________________

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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19 Jun 2014, 01:06

If x, y, and z are positive integers and xyz=2,700. Is x√ an integer? (1) y is an even perfect square and z is an odd perfect cube.

(2) z√ is not an integer.

For this question, I agree one of the values can be 1. But issue is we dont apply this logic to all questions. Somewhere we do this and sometimes we don't. I did a question before in which 1 was not considered. This is really a trap.

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 Jul 2014, 10:20

Bunuel wrote:

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) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.

Hi Bunuel,

Could you please provide me some explanation for (2) when you say " since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12"?

How can we assume that x - y = 1? It could have been any two numbers. Is there a theory that if difference of square of two numbers, then they are consecutive? Could you please provide some reference?

Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 Jul 2014, 10:33

Expert's post

gkashyap wrote:

Bunuel wrote:

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) \(x^2-y^2=23\)

(1) \(100<y^2<x^2<169\) --> since both \(x\) and \(y\) are positive integers then \(x^2\) and \(y^2\) are perfect squares --> there are only two perfect squares in the given range 121=11^2 and 144=12^2 --> \(y=11\) and \(x=12\). Sufficient.(As cyclicity of units digit of \(7\) in integer power is \(4\), therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), so 3).

(2) \(x^2-y^2=23\) --> \((x-y)(x+y)=23=prime\) --> since both \(x\) and \(y\) are positive integers then: \(x-y=1\) and \(x+y=23\) --> \(y=11\) and \(x=12\). Sufficient.

Answer: D.

Hi Bunuel,

Could you please provide me some explanation for (2) when you say " since both x and y are positive integers then: x-y=1 and x+y=23 --> y=11 and x=12"?

How can we assume that x - y = 1? It could have been any two numbers. Is there a theory that if difference of square of two numbers, then they are consecutive? Could you please provide some reference?

Thanks for the help.

Regards, Gajendra

\((x-y)(x+y)=23=prime\). 23 is a prime number, so it can be broken into the product of two positive multiples only in one way 23=1*23. Now, since x and y are positive integers, then x-y<x+y, thus x-y=1 and x+y=23.

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