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# NEW!!! Tough and tricky exponents and roots questions

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

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Updated on: 15 Mar 2019, 04:10
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395
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. 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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029216

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029219

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. 14/5
B. 5
C. 28/5
D. 13
E. 14

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029221

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029222

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029223

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029224

7. If $$x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}$$, then which of the following must be true:
A. x<6
B. 6<x<8
C. 8<x<10
D. 10<x<12
E. x>12

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029227

8. If $$x$$ is a positive number and equals 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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029228

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. $$2^{22x}*11^{22x}$$

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029229

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029231

11. If $$x>0$$, $$x^2=2^{64}$$ and $$x^x=2^y$$ then what is the value of $$y$$?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029232
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Originally posted by Bunuel on 12 Jan 2012, 03:03.
Last edited by Bunuel on 15 Mar 2019, 04:10, edited 1 time in total.
Updated.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Jan 2012, 14:53
28
50
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

Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know fro the GMAT: $$(x+y)^2=x^2+2xy+y^2$$ (while $$(x-y)^2=x^2-2xy+y^2$$).

So 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 sum of the first and the third terms simplifies to $$(25+10\sqrt{6})+(25-10\sqrt{6})=50$$, so we have $$50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})$$ --> $$50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}$$.

Also must know for the GMAT: $$(x+y)(x-y)=x^2-y^2$$, thus $$50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)}=50+2\sqrt{625-600}=50+2\sqrt{25}=60$$.

Recall that we should un-square this value to get the right the answer: $$\sqrt{60}=2\sqrt{15}$$.

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

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13 Jan 2012, 21:46
3
8
Q4. What is the value of $$5+4*5+4*5^2+4*5^3+4*5^4+4*5^5$$

This one was actually the simplest I thought. Here is how:

$$5+4*5+4*5^2+4*5^3+4*5^4+4*5^5$$

So $$5*(1+4)+4*5^2+4*5^3+4*5^4+4*5^5$$
So $$5^2+4*5^2+4*5^3+4*5^4+4*5^5$$
So $$5^2*(1+4)+4*5^3+4*5^4+4*5^5$$
So $$5^3+4*5^3+4*5^4+4*5^5$$
So $$5^3*(1+4)+4*5^4+4*5^5$$

So every expression behind contributes a power of 1 to the one in front of it. We just need to see the last which is $$5^5$$
Keep solving and you come to a total of $$5^6$$

Hence Answer = $$5^6$$
##### General Discussion
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 03:10
4
1
8
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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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 03:43
1
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

( \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2 =60

\sqrt{60}=2\sqrt{15}
ans is
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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Updated on: 12 Jan 2012, 12:26
2. What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8

7^12-3^9
11-3=8

ANS IS E
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Originally posted by LalaB on 12 Jan 2012, 04:04.
Last edited by LalaB on 12 Jan 2012, 12:26, edited 1 time in total.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 04:34
imho 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. 14/5
B. 5
C. 28/5
D. 13
E. 14
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 04:56
LalaB wrote:
2. What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8

7^12-3^9
1-1=0

ANS IS A

I get 2. for 1st term 1 and for the 2nd term 9. 1st term - 2nd term= unit digit=2
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 05:06
1
Q1= root (a+b)^2 [say: \sqrt{25+10\sqrt{6}}= a and \sqrt{25-10\sqrt{6}}= b ]
= root (50+2root(25))
= root(60)
= 2root(15)
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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Updated on: 12 Jan 2012, 05:32
1
Edited
What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
= 5+4*5[1+5+5^2+5^3+5^4]
=5+20*781
=15625
=625*25=5^6

Originally posted by BDSunDevil on 12 Jan 2012, 05:19.
Last edited by BDSunDevil on 12 Jan 2012, 05:32, edited 1 time in total.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 05:23
1
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

5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5*5+4*5^2(1+5)+4*5^4(1+5)=
5^2+4*5^2*6+4*5^4*6=5^4+4*5^4*6=5^4*5^2=5^6
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 09:45
for 1:
I just read it as (a + b)
i. e = \sqrt{(a+b)^2}
then substituting the value for a and b
a= \sqrt{(25+10[square_root]6)}[/square_root] and b = a= \sqrt{(25-10[square_root]6)}[/square_root]
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 09:49
1
wallstreetbarbie wrote:
Can you explain 1? some of the code came out messed up and its difficult to read

\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}} )

1st raise it to the 2nd power (in a simple form u have (a+b)^2)
(\sqrt{25+ 10[square_root]6}[/square_root]+\sqrt{25- 10[square_root]6}[/square_root] )^2 =

=((\sqrt{25+ 10[square_root]6}[/square_root])^2+ 2((\sqrt{25+ 10[square_root]6}[/square_root])*(\sqrt{25- 10[square_root]6}[/square_root] ) +(\sqrt{25- 10[square_root]6}[/square_root] )^2 (in a simple form u have got (a^2+2ab+b^2)

=25+ 10\sqrt{6}[/square_root]+25-10\sqrt{6}[/square_root] +2((\sqrt{25+ 10[square_root]6}[/square_root])*(\sqrt{25- 10[square_root]6}[/square_root]))

=50+2*\sqrt{625-6*100}=50+2*5=60

\sqrt{60}=2\sqrt{15}

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

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12 Jan 2012, 10:10
2
8- B
the entire expression $$\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}$$
we can write as
x = $$\sqrt{6+x}$$ ----as its an infinite series

so $$x^2$$ = 6+x
$$x^2$$ -x - 6 = 0

(x-3)(x+2) =0

as x is +ve x=3
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 10:12
1
Q 5:

2^n*13^n * K= 23^2*5^8*3^18*29^8 [ after prime factorization]
value of n has to be 0 for 2^n*13^n to be 1
therefore, n^2*6 -26^n
=0-1
=-1
(??)
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 10:44
2
Q 10. from equation 1 and 2: we can solve for x= 45
we also know: z-3y= 5x - 125

substituting these: we get the value= 10
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 11:13
1
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

23^2*25^4*27^6*29^8/26^n= 23^2*25^4*27^6*29^8/(2*13)^n from this point it is obvious, that n =0 (since both 2 and 13 are primes and none of the numbers of numerator can be divisible by 2 and 13)

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

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

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12 Jan 2012, 11:23
1
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

hm, have doubts in my solution...

x=\sqrt[5]{-37}

37 is near 32 . 32 =2^5 it means that -3<x<-2

so D is correct
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 11:48
11. If x>0, x^2=2^{64} and x^x=2^y then what is the value of y?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)

x^2=2^{64}
x^2=(2^{32})^2 so x =2^32 from now I dont know ...either C, or E is the answ. I would choose C

@Bunuel, when r u going to post answers? just curious
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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12 Jan 2012, 12:18
2
1
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
Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 12 Jan 2012, 12:18

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