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3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\), what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}\)?

A. \(\frac{14}{5}\)
B. 5
C. \(\frac{28}{5}\)
D. 13
E. 14


The first thing one should notice here is that \(x\) and \(y\) must be some irrational numbers. This is because 4,900 has primes other than 5 in its prime factorization, and 25 doesn't have 2 as a prime at all. Thus, we should manipulate the given expressions rather than solve for \(x\) and \(y\) directly.

\(5^{10x}=4,900\)

\((5^{5x})^2=70^2\)

\(5^{5x}=70\)

Now, \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}} = 5^{5x}*5^{-5}*(2^{\sqrt{y}})^2\)

Since \(5^{5x}=70\), then \(5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.
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10. Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

By rearranging both expressions, we get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Let's denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\), then \((a-b)(a+b)=5*(a+b)=125\), which implies that \(a+b=25\).

This gives us two equations with two unknowns: \(a+b=25\) and \(a-b=5\). Solving for \(a\) yields \(a=15=\sqrt{5x}\), so \(x=45\). Solving for \(b\) yields \(b=10=\sqrt{z-3y}\).

Finally, \(\sqrt{\frac{45(z-3y)}{x} }=\sqrt{\frac{45*10^2}{45} }=10\).


Answer: B
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2. What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


MUST KNOW FOR THE GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\).

    Hence, the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. \({(a^m)}^n=a^{mn}\) and \(a^{m^n}=a^{(m^n)}\).

    Hence, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer powers repeats in a specific pattern called cyclicity. The units digit of 7 and 3 in positive integer powers repeats in patterns of 4:

    1. \(7^1=7\)(last digit is 7)

    2. \(7^2=49\)(last digit is 9)

    3. \(7^3=xx3\)(last digit is 3)

    4. \(7^4=xxx1\) (last digit is 1)

    5. \(7^5=xxxxxx7\) (last digit is 7 again!)

    ...

    1. \(3^1=3\) (last digit is 3)

    2. \(3^2=9\) (last digit is 9)

    3. \(3^3=27\) (last digit is 7)

    4. \(3^4=81\) (last digit is 1)

    5. \(3^5=243\) (last digit is 3 again!)

    ...

Hence, the units digit of \(17^{12}\) is 1, as the 12th number in the repeating pattern 7-9-3-1 is 1, and the units digit of \(1973^9\) is 3, as the 9th number in the repeating pattern 3-9-7-1 is 3.

BACK TO THE QUESTION:

Thus, we know that the units digit of \({(17^3)}^4=17^{12}\) is 1, and the units digit of \(1973^{3^2}=1973^9\) is 3. If the first number is greater than the second number, the units digit of their difference will be 8, for example, 11 - 3 = 8. However, if the second number is greater than the first number, the units digit of their difference will be 2, for example, 11 - 13 = -2.

Consider this, even if the first number were \(100^{12}\) instead of \(17^{12}\), and the second number were \(1000^9\) instead of \(1973^9\), the first number, \(100^{12}=10^{24}\), would still be smaller than the second number \(1,000^9=10^{27}\). Therefore, \(17^{12} < 1973^9\), and the units digit of the difference is 2.

Answer B.
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8. If \(x\) is a positive number and is equal to \(\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...} } } } }\), where the given expression extends to an infinite number of roots, then what is the value of \(x\)?

A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6

Given: \(x \gt 0\) and \(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...} } } } }\).

Re-write it as: \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...} ) } } } }\). As the expression under the square root extends infinitely, the expression in brackets would equal to \(x\) itself. Therefore, we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\). Square both sides to get: \(x^2=6+x\). Now, re-arrange and factorize to obtain \((x+2)(x-3)=0\). This gives us \(x=-2\) or \(x=3\). But since \(x \gt 0\), the valid solution is: \(x=3\).

Answer: B.
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SOLUTIONS:

1. What is the value of \( \sqrt{25 + 10 \sqrt{6} } + \sqrt{ 25 - 10 \sqrt{6} }\) ?

A. \(2\sqrt{5}\)
B. \(\sqrt{55}\)
C. \(2\sqrt{15}\)
D. 50
E. 60

First, square the given expression to eliminate the square roots, but remember to take the square root of the result at the end to balance the operation and obtain the correct answer.

Important for the GMAT: \((x+y)^2=x^2+2xy+y^2\) and \((x-y)^2=x^2-2xy+y^2\).

Following these rules, we get:

    \((\sqrt{25 + 10 \sqrt{6} } + \sqrt{ 25 - 10 \sqrt{6} })^2 =\)

    \(=(\sqrt{25 + 10\sqrt{6} })^2+2(\sqrt{25 + 10\sqrt{6} })(\sqrt{25 - 10\sqrt{6} })+(\sqrt{25 - 10\sqrt{6} })^2=\)

    \(=(25+10\sqrt{6})+2(\sqrt{25 + 10\sqrt{6} })(\sqrt{25 - 10\sqrt{6} })+(25-10\sqrt{6})\).

Note that the sum of the first and third terms simplifies to \((25+10\sqrt{6})+(25-10\sqrt{6})=50\), so we have:

    \(50+2(\sqrt{25 + 10\sqrt{6} })(\sqrt{25 - 10\sqrt{6} })=\)

    \(=50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6}) }\).


Another important concept for the GMAT: \((x+y)(x-y)=x^2-y^2\). Using this, we can simplify further:

    \(50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})}=\)

    \(=50+2\sqrt{25^2-(10\sqrt{6})^2)} = \)

    \( = 50+2\sqrt{625-600}=\)

    \(=50+2\sqrt{25}=\)

    \(=60\).

Finally, remember to take the square root of this value to obtain the correct answer: \(\sqrt{60}=2\sqrt{15}\).

Answer: C.
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4. What is the value of \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5\)?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10

This question can be solved in several ways:

Traditional approach:

\(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5)\). Note that we have the sum of a geometric progression in brackets with the first term equal to 5 and the common ratio also equal to 5. The sum of the first \(n\) terms of a geometric progression is given by: \(\text{sum}=\frac{b*(r^{n}-1)}{r-1}\), where \(b\) is the first term, \(n\) is the number of terms, and \(r\) is a common ratio \(\ne 1\).

For our equation: \(5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6\).

30 sec approach based on answer choices:

We have the sum of 6 terms. Now, if all terms were equal to the largest term \(4*5^5\), we would have: \(\text{sum}=6*(4*5^5)=24*5^5\approx 5^2*5^5 \approx 5^7\), so the actual sum must be less than \(5^7\), thus the answer must be A: \(5^6\).

Answer: A.
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22. If \(x\) is a positive integer, which of the following is closest to \(\frac{22^{22x} - 22^{2x} }{11^{11x} - 11^x}\)?

A. \(2^{11x}\)
B. \(11^{11x}\)
C. \(22^{11x}\)
D. \(2^{22x}*11^{11x}\)
E. \(2^{22x}*11^{22x}\)

First of all, it's important to note that we are looking for an approximate value of the given expression.

Now, considering that \(22^{22x}\) is significantly larger than \(22^{2x}\), the value of \(22^{22x}-22^{2x}\) will be very close to just \(22^{22x}\). In this context, \(22^{2x}\) becomes negligible. Similarly, \(11^{11x}-11^x\) will be almost equivalent to \(11^{11x}\)

Therefore, \(\frac{22^{22x}-22^{2x} }{11^{11x}-11^x} \approx \frac{22^{22x} }{11^{11x} } = \frac{2^{22x}*11^{22x} }{11^{11x} } = 2^{22x}*11^{11x}\).

For further verification, one can also approach this algebraically: \(\frac{22^{22x}-22^{2x} }{11^{11x}-11^x} = \frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\). Since the -1 in both the denominator and numerator is negligible, the expression simplifies to: \(\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)} \approx \frac{22^{2x}*22^{20x} }{11^x*11^{10x} } = \frac{22^{22x} }{11^{11x} } = 2^{22x}*11^{11x}\).

Answer: D.
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6. If \(x=\sqrt[5]{-37}\) then which of the following must be true?
A. \(\sqrt{-x}>2\)
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32

MUST KNOW FOR THE GMAT:

• Even roots from a negative number are undefined on the GMAT (as GMAT is dealing only with Real Numbers): \(\sqrt[{even}]{negative}=undefined\), for example, \(\sqrt{-25}=undefined\).

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

BACK TO THE ORIGINAL QUESTION:

As \(-2^5=-32\), then \(x\) must be a little bit less than -2, hence \(x=\sqrt[5]{-37} \approx -2.1 \lt -2\). Thus \(x^3 \approx (-2.1)^3 \approx -8.something \lt -8\), so option D must be true.

As for the other options:

A. \(\sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1} \lt 2\), \(\sqrt{-x} \gt 2\) is not true.

B. \(x \approx -2.1 \lt -2\), thus \(x \gt -2\) is also not true.

C. \(x^2 \approx (-2.1)^2=4.something \gt 4\), thus \(x^2 \lt 4\) is also not true.

Answer: D.
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11. If \(x \gt 0\), \(x^2=2^{64}\), and \(x^x=2^y\), what is the value of \(y\)?

A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)

Given that \(x^2 = 2^{64}\), we can deduce that \(x = \sqrt{2^{64} } = 2^{\frac{64}{2} } = 2^{32}\). (Note: The solution \(x = -\sqrt{2^{64} }\) is not valid because we are given that \(x > 0\)).

For the second step: \(x^x = (2^{32})^{(2^{32})} = 2^{32*2^{32} } = 2^{2^{5}*2^{32} } = 2^{2^{37} } = 2^y\). So, we have that \(2^{2^{37} } = 2^y\), which means that \(y = 2^{37}\).

Alternatively: \(x^x = (2^{32})^x = 2^{32x} = 2^y\). So, we have that \(2^{32x} = 2^y\), which implies \(y = 32x\). Given that \(x = 2^{32}\), then \(y = 32 * 2^{32} = 2^5 * 2^{32} = 2^{37}\).

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. What is the value of \sqrt{(25+10√6)} + \sqrt{(25-10√6)}?
A. 2√5
B. √55
C. 2√15
D. 50
E. 60


N = √(25+10√6) + √(25-10√6)
N^2 = 25+10√6 + 25-10√6 + 2\sqrt{625-600}
N^2 = 50 + 2*5 = 60
N = √60 = 2√15

Option
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2. What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8


Units digit for 17^3 = 3
Units digit for (17^3)^4 = 3^4 = 1
Units digit for 1973^9 = 3
Units digit for xxx1 - xxxx3 = 8

Option
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5. If x=23^2*25^4*27^6*29^8 and is a multiple of 26^n, where n is a non-negative integer, then what is the value of n^26-26^n?
A. -26
B. -25
C. -1
D. 0
E. 1

n=0 because x is not a multiple of 26

n^26 - 26^n = 0-1 = -1

Option
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6. If x=\sqrt[5]{-37} then which of the following must be true?
A. \sqrt{-x}>2
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32


x^5 = -37

-2^5 = -32
-3<x<-2

A. Eliminated
B. Eliminated
C. x^2<4 = -2<x<2 Eliminated
D. x^3<-8 Correct


Option
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
9. If x is a positive integer then the value of \frac{22^{22x}-22^{2x}}{11^{11x}-11^x} is closest to which of the following?
A. 2^{11x}
B. 11^{11x}
C. 22^{11x}
D. 2^{22x}*11^{11x}
E. 2^{22x}*11^{22x}


let 11=a

so we have

(2a^(2ax)-2a^2x)/(a^(ax)-a^x)=2*((a^(ax)-a^x)*(a^(ax)+a^x))/(a^(ax)-a^x)=2(a^(ax)+a^x)

2(a^(ax)+a^x)=2*(11^(11x)+11^x)
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LalaB wrote:
9. If x is a positive integer then the value of \frac{22^{22x}-22^{2x}}{11^{11x}-11^x} is closest to which of the following?
A. 2^{11x}
B. 11^{11x}
C. 22^{11x}
D. 2^{22x}*11^{11x}
E. 2^{22x}*11^{22x}


let 11=a

so we have

(2a^(2ax)-2a^2x)/(a^(ax)-a^x)=2*((a^(ax)-a^x)*(a^(ax)+a^x))/(a^(ax)-a^x)=2(a^(ax)+a^x)

2(a^(ax)+a^x)=2*(11^(11x)+11^x)


And it looks as though every other question has been solved except for this one. I won't solve it completely, but I'll help people get started. Here, since we are asked for an approximation only, we don't need to compute an exact value. You might notice that if x is a positive integer, \(22^{22x}\) is vastly bigger than \(22^{2x}\). Since that's true, if we only need to estimate, we can ignore the small term. We can do the same thing in the denominator.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
IanStewart wrote:

And it looks as though every other question has been solved except for this one. I won't solve it completely, but I'll help people get started. Here, since we are asked for an approximation only, we don't need to compute an exact value. You might notice that if x is a positive integer, \(22^{22x}\) is vastly bigger than \(22^{2x}\). Since that's true, if we only need to estimate, we can ignore the small term. We can do the same thing in the denominator.


2a^(2ax)/a^(ax)=2*a^(ax)=2*11^(11x)

ans is ?
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