NEW!!! Tough and tricky exponents and roots questions

2. If \(x\), \(y\), and \(z\) are positive integers and \(xyz=2,700\). Is \(\sqrt{x}\) an integer?

Note: a perfect square is an integer that can be written as the square of an 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 an integer. For example, \(27=3^3\) is a perfect cube.

First, find the prime factorization of 2,700: \(xyz=2,700=2^2*3^3*5^2\).

Now, let's examine the given statements:

(1) \(y\) is an even perfect square and \(z\) is an odd perfect cube.

Since \(z\) can take fewer values than \(y\), for simplicity, let's analyze it from \(z\) perspective. Since given that \(z\) is an odd perfect cube, it can be \(1=1^3\) or \(3^3\).

If \(z = 3^3\), \(y\) can only be \(2^2\) or \(2^2*5^2\). Respectively, \(x\) will be \(5^2\) or \(1\). For each of these values, \(\sqrt{x}\) is an integer.

If \(z = 1\), \(y\) can be \(2^2\), \(2^2*3^2\), \(2^2*5^2\), or \(2^2*3^2*5^2\). For the first three cases, \(x\) turns out NOT to be a perfect square (\(3^3*5^2\), \(3*5^2\), and \(3^3\)). However, for the fourth case, \(\sqrt{x}\) is a perfect square (\(1\)).

Not sufficient.

(2) \(\sqrt{z}\) is not an integer. This statement is clearly insufficient, as it does not provide enough information about \(x\) and \(y\).

(1)+(2) Since from (2) \(\sqrt{z} \ne integer\), then \(z \ne 1\), and thus from (1) it can only be \(3^3\). As discussed above, if \(z=3^3\), then \(x\) will be \(5^2\) or \(1\). For each of these values, \(\sqrt{x}\) is an integer. Sufficient.


Answer: C
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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.
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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) \(x-y=9\)

\(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\) --> factor out \(\sqrt{2}\) from the nominator and apply \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator: \(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). So we should find the value of \(\sqrt{x}-\sqrt{y}\).

(1) \(x+y=4+2\sqrt{xy}\) --> \(x-2\sqrt{xy}+y=4\) --> \((\sqrt{x}-\sqrt{y})^2=4\) --> \(\sqrt{x}-\sqrt{y}=2\) (note that since \(x-y>0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.

(2) \(x-y=9\). Not sufficient.

Answer: A.
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1. If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)?

(1) \(100 \lt y^2 \lt x^2 \lt 169\).

Since both \(x\) and \(y\) are positive integers, \(x^2\) and \(y^2\) are perfect squares. Now, there are only two perfect squares in the given range: \(121=11^2\) and \(144=12^2\), so \(y=11\) and \(x=12\). Sufficient. (As the cyclicity of the units digit of 7 in integer power is 4, therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), which is 3).

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

From this equation, we have \((x-y)(x+y) = 23\). Since 23 is prime and both \(x\) and \(y\) are positive integers, it follows that \(x-y=1\) and \(x+y=23\). Solving for \(x\) and \(y\), we find \(y=11\) and \(x=12\). Sufficient.

Answer: D.
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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.

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

(1) \(2^{x-y}=\sqrt[(x+y)]{16}\) --> \(2^{x-y}=16^{\frac{1}{x+y}}=2^{\frac{4}{x+y}}\) --> equate the powers: \(x-y=\frac{4}{x+y}\) --> \((x-y)(x+y)=4\).

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

Two different answers. Not sufficient.

Answer: A.
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8. What is the value of \(x^2+y^3\)?
(1) \(x^6+y^9=0\)
(2) \(27^{x^2}=\frac{3}{3^{3y^2+1}}\)

(1) \(x^6+y^9=0\) --> \((x^2)^3=(-y3)^3\) ---> \(x^2=-y^3\) --> \(x^2+y^3=0\). 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.

Answer: D.
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5. If \(x\) and \(y\) are negative integers, then what is the value of \(xy\)?
(1) \(x^y=\frac{1}{81}\)
(2) \(y^x=-\frac{1}{64}\)

(1) \(x^y=\frac{1}{81}\) --> as both \(x\) and \(y\) are negative integers then \(x^y=\frac{1}{81}=(-9)^{-2}=(-3)^{-4}\) --> \(xy=18\) or \(xy=12\). Note that as negative integer (x) in negative integer power (y) gives positive number (1/81) then the power must be negative even number. Not sufficient.

(2) \(y^x=-\frac{1}{64}\) --> as the result is negative then \(x\) must be negative odd number --> \(y^x=-\frac{1}{64}=(-4)^{-3}=(-64)^{-1}\) --> \(xy=12\) or \(xy=64\). Not sufficient.

(1)+(2) Only one pair of negative integers \(x\) and \(y\) satisfies both statements \(x=-3\) and \(y=-4\) --> \(xy=12\). Sufficient.

Answer: C.
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4. If \(xyz\neq{0}\) is \((x^{-4})*(\sqrt[3]{y})*(z^{-2})<0\)?
(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\)
(2) \(y^3>\frac{1}{z{^4}}\)

\(xyz\neq{0}\) means that neither of unknown is equal to zero. Next, \((x^{-4})*(\sqrt[3]{y})*(z^{-2})=\frac{\sqrt[3]{y}}{x^4*z^2}\), so the question becomes: is \(\frac{\sqrt[3]{y}}{x^4*z^2}<0\)? Since \(x^4\) and \(z^2\) are positive numbers then the question boils down whether \(\sqrt[3]{y}<0\), which is the same as whether \(y<0\) (recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\)).

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}\) --> as even root from positive number (\(x^2\) in our case) is positive then \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

(2) \(y^3>\frac{1}{z{^4}}\) --> the same here as \(\frac{1}{z{^4}}>0\) then \(y^3>\frac{1}{z{^4}}>0\), (or which is the same \(y>0\)). Therefore answer to the original question is NO. Sufficient.

Answer: D.
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6. If \(x>0\) then what is the value of \(y^x\)?
(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\)
(2) \(x\neq{1}\) and \(x^y=1\)

(1) \(\frac{4^{(x+y)^2}}{4^{(x-y)^2}}=128^{xy}\) --> \(4^{(x+y)^2-(x-y)^2}=128^{xy}\) --> applying \(a^2-b^2=(a-b)(a+b)\) we'll get: \(4^{4xy}=128^{xy}\) --> \(2^{8xy}=2^{7xy}\) --> \(8xy=7xy\) --> \(xy=0\), since given that \(x>0\) then \(y=0\) hence \(y^x=0^x=0\). Sufficient.

(2) \(x\neq{1}\) and \(x^y=1\) --> since \(x>0\) and \(x\neq{1}\) then the only case \(x^y=1\) to hold true is when \(y=0\) --> \(y^x=0^x=0\). Sufficient.

Answer: D.
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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.

Answer: D.
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THEORY TO TACKLE THE PROBLEMS ABOVE:
For more on number theory check the Number Theory Chapter of Math Book: math-number-theory-88376.html

EXPONENTS

Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number \(a\) multiplied \(n\) times can be written as \(a^n\), where \(a\) represents the base, the number that is multiplied by itself \(n\) times and \(n\) represents the exponent. The exponent indicates how many times to multiple the base, \(a\), by itself.

Exponents one and zero:
\(a^0=1\) Any nonzero number to the power of 0 is 1.
For example: \(5^0=1\) and \((-3)^0=1\)
• Note: the case of 0^0 is not tested on the GMAT.

\(a^1=a\) Any number to the power 1 is itself.

Powers of zero:
If the exponent is positive, the power of zero is zero: \(0^n = 0\), where \(n > 0\).

If the exponent is negative, the power of zero (\(0^n\), where \(n < 0\)) is undefined, because division by zero is implied.

Powers of one:
\(1^n=1\) The integer powers of one are one.

Negative powers:
\(a^{-n}=\frac{1}{a^n}\)

Powers of minus one:
If n is an even integer, then \((-1)^n=1\).

If n is an odd integer, then \((-1)^n =-1\).

Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
\(a^n*b^n=(ab)^n\)

\(\frac{a^n}{b^n}=(\frac{a}{b})^n\)

\((a^m)^n=a^{mn}\)

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down)

Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
\(a^n*a^m=a^{n+m}\)

\(\frac{a^n}{a^m}=a^{n-m}\)

Fraction as power:
\(a^{\frac{1}{n}}=\sqrt[n]{a}\)

\(a^{\frac{m}{n}}=\sqrt[n]{a^m}\)


ROOTS

Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.

General rules:
• \(\sqrt{x}\sqrt{y}=\sqrt{xy}\) and \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\).

• \((\sqrt{x})^n=\sqrt{x^n}\)

• \(x^{\frac{1}{n}}=\sqrt[n]{x}\)

• \(x^{\frac{n}{m}}=\sqrt[m]{x^n}\)

• \({\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}\)

• \(\sqrt{x^2}=|x|\), when \(x\leq{0}\), then \(\sqrt{x^2}=-x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\)

• 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, +5 and -5. Even roots have only a positive value on the GMAT.

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

10<y<x<13
y = 11
x = 12

Sufficient

(2) x^2-y^2=23
(x+y)(x-y) = 23
x-y=1
x+y=23
x=12
y=11

Sufficient

Option

if we can find the sign of y we can answer the question.

(1) \(\sqrt[5]{y}>\sqrt[4]{x^2}>0\) since it is\([\sqrt[4]{x^2}]=[\sqrt[]{x}]\)

Sufficient

(2) \(y^3>\frac{1}{z{^4}}>0\) since \(z^4>0\)

Sufficient

Therefore, D

(1) y must be even for the statement to be true (only then will the result be positive). \(1/81=\) 3^(-4) ---> the pairs to make 4 in order for x and y to be integers is 1*4 or 2*2. Therefore the only way to construct it is if y=-2 or y=-4.

if x=-3 and y=-4 --->\(x^y=\frac{1}{81}\) ---> x*y=12
if x=-9 and y=-2 --->\(x^y=\frac{1}{81}\) ---->x*y=18

Insufficient

(2) x must be odd for the statement to be True (only then will the result be negative).\(-1/64=-2^6\) ---> the pairs to make 6 in order for x and y to be integers are 1*6 or 2*3. Therefore the only way to construct it is if x=-1 or x=-3.

if y=-4 and x=-3 ---> \(y^x=-\frac{1}{64}\) --->x*y=12
if y=-64 and x=-1 ---> \(y^x=-\frac{1}{64}\) ---> x*y=64

Insufficient.

(1)+(2)---> x*y=12 ---> C

(1) x must have a 7 raised at an odd power as a factor. It could also have any number raised to an even power as a factor ---> therefore \(\sqrt{x}\) will never be an integer because of the 7. Sufficient

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

Sufficient

Therefore, D

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,

\(x - y = (\sqrt{x})^2 - (\sqrt{y})^2 = (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\)

So we can simplify the question by using this factorization in the denominator:

\(\frac{\sqrt{2x} + \sqrt{2y}}{x-y} = \frac{\sqrt{2} (\sqrt{x} + \sqrt{y})}{(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) } = \frac{\sqrt{2}}{\sqrt{x} - \sqrt{y}}\)

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.
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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.
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