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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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12 Jan 2012, 02:03
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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. 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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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25 Jun 2012, 08:59
kuttingchai wrote:
Bunuel wrote:
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 $$c^n$$, which means that the units digit of $$(17^3)^4$$ is that same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is that same as that of $$3^{3^2}$$.

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$, which on the other hand equals to $$a^{mn}$$.

So:
$$(a^m)^n=a^{mn}$$;

$$a^m^n=a^{(m^n)}$$.

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

III. The units digit of integers in positive integer power repeats in specific pattern (cyclicity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (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!)
...

Thus th units digit of $$7^{12}$$ will be 1 (4th in pattern, as 12 is a multiple of cyclicty number 4) and the units digit of $$3^9$$ will be 3 (first in pattern, as 9=4*2+1).

So, we have that the units digit of $$(17^3)^4=17^{12}$$ is 1 and the units digit of $$1973^3^2=1973^9$$ is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of $$(17^3)^4-1973^{3^2}$$ is 2.

Hey Bunuel,

I understood the logic behind finding the unit places, but how can u determine if the reminder is 2 or 8

in this example
(17^12) - (1973^9) = unit place is 2 (agreed- i actually did the calculation using calculator )
because : (17^12) unit place = 1
(1973^9) unit place = 3

but if we have just - (7^12) - (3^9) or (3^9) - (7^12) we have unit place as "8"

3^9 = 19683 - unit place will still be 2
7^12 = 13841287201 - unit place will still be 1

so my question is how will u determine if the answer is 2/8? clearly we have 2 different answers??
am i missing anything??

The units digit of $$17^{12}-1973^{9}$$ is 2 and not 8 since $$1973^{9}$$ is much larger number than $$17^{12}$$, thus their difference will be negative, something like 11-13=-2.

If we had something like 21-13 (if the first number were greater than the second one), then the units digit of their difference would be 8.

Hope i's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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29 Jun 2012, 10:34
Bunuel wrote:
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: $$\sqrt{60}=2\sqrt{15}$$.

Another way to do it, using the same formulas:

$$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{5(5+2\sqrt{6})}+\sqrt{5(5-2\sqrt{6})}=\sqrt{5}\sqrt{5+2\sqrt{6}}+\sqrt{5}\sqrt{5-2\sqrt{6}}=$$
$$=\sqrt{5}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=\sqrt{5}(\sqrt{(\sqrt{3}+\sqrt{2})^2}+\sqrt{(\sqrt{3}-\sqrt{2})^2})=\sqrt{5}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=\sqrt{5}(2\sqrt{3})=2\sqrt{15}$$
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Last edited by EvaJager on 15 Oct 2012, 00:35, edited 1 time in total.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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25 Jul 2012, 12:53
Bunuel wrote:
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: $$\sqrt{60}=2\sqrt{15}$$.

I don't understand why we should un-square the value at the end?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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26 Jul 2012, 05:10
Stiv wrote:
Bunuel wrote:
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: $$\sqrt{60}=2\sqrt{15}$$.

I don't understand why we should un-square the value at the end?

Please read the solution: "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."
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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27 Aug 2012, 15:58
These questions are absolutely insane, do people really solve these on the GMAT under 2 or 3 minutes? I just recently took a GMATprep practice test and got a 690 (a measly 44Q) and nowhere did I see any problems of this caliber. Is it truly what people deal with to get to 46 or 47 Quant?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 Sep 2012, 02:29
Bunuel wrote:
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)

$$x^2=2^(64)$$ --> $$x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32}$$ (note that $$x=-\sqrt{2^{64}}$$ is not a valid solution as given that $$x>0$$).

Second step: $$x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y$$ --> $$y=2^{37}$$.

OR second step: $$x^x=(2^{32})^x=2^{32x}=2^y$$ --> $$y=32x$$ --> since $$x=2^{32}$$ then $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

Bunuel I have little confusion about the last problem

given $$x^2$$= $$2^{64}$$so we can write $$2^{64}$$ as $$(2^{32})^{2} = x^2$$ now we have $$x = 2^{32}$$

now second part $$x^x = {2^{32}}^ {x}$$ (i)

or does it mean

$$x^x= 2^{{32}{x}}$$ ( ii)

in (i) we have $$32^x$$and in (ii) we have $$32*x$$ obviously they cannot mean the same thing $${{2^{32}}^2}^{32} \neq{2^{32.2}}^{32}$$ ,as $${2^{32}}^x \neq 2^{32.x}$$ $$(x>1)$$

in the solution you have taken $${2^{32}}^{x} = 2^{{32}{x}}$$ so I am confused

so we are saying $${{2^{32}}^2}^{32} = {2^{32.2}}^{32}$$..how ?

sorry for the silly question, ..cannot understand how $$32^x = 32.x$$ please help , again sorry
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Sep 2012, 00:57
stne wrote:
Bunuel wrote:
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)

$$x^2=2^(64)$$ --> $$x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32}$$ (note that $$x=-\sqrt{2^{64}}$$ is not a valid solution as given that $$x>0$$).

Second step: $$x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y$$ --> $$y=2^{37}$$.

OR second step: $$x^x=(2^{32})^x=2^{32x}=2^y$$ --> $$y=32x$$ --> since $$x=2^{32}$$ then $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

Bunuel I have little confusion about the last problem

given $$x^2$$= $$2^{64}$$so we can write $$2^{64}$$ as $$(2^{32})^{2} = x^2$$ now we have $$x = 2^{32}$$

now second part $$x^x = {2^{32}}^ {x}$$ (i)

or does it mean

$$x^x= 2^{{32}{x}}$$ ( ii)

in (i) we have $$32^x$$and in (ii) we have $$32*x$$ obviously they cannot mean the same thing $${{2^{32}}^2}^{32} \neq{2^{32.2}}^{32}$$ ,as $${2^{32}}^x \neq 2^{32.x}$$ $$(x>1)$$

in the solution you have taken $${2^{32}}^{x} = 2^{{32}{x}}$$ so I am confused

so we are saying $${{2^{32}}^2}^{32} = {2^{32.2}}^{32}$$..how ?

sorry for the silly question, ..cannot understand how $$32^x = 32.x$$ please help , again sorry

Bunuel ( preferably ) or any one else , can you please provide solution to the above query , This question was given by you
seems something is odd with the solution as pointed in my post above, unless I am missing something .
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Sep 2012, 01:07
stne wrote:
Bunuel wrote:
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)

$$x^2=2^(64)$$ --> $$x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32}$$ (note that $$x=-\sqrt{2^{64}}$$ is not a valid solution as given that $$x>0$$).

Second step: $$x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y$$ --> $$y=2^{37}$$.

OR second step: $$x^x=(2^{32})^x=2^{32x}=2^y$$ --> $$y=32x$$ --> since $$x=2^{32}$$ then $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

Bunuel I have little confusion about the last problem

given $$x^2$$= $$2^{64}$$so we can write $$2^{64}$$ as $$(2^{32})^{2} = x^2$$ now we have $$x = 2^{32}$$

now second part $$x^x = {2^{32}}^ {x}$$ (i)

or does it mean

$$x^x= 2^{{32}{x}}$$ ( ii)

in (i) we have $$32^x$$and in (ii) we have $$32*x$$ obviously they cannot mean the same thing $${{2^{32}}^2}^{32} \neq{2^{32.2}}^{32}$$ ,as $${2^{32}}^x \neq 2^{32.x}$$ $$(x>1)$$

in the solution you have taken $${2^{32}}^{x} = 2^{{32}{x}}$$ so I am confused

so we are saying $${{2^{32}}^2}^{32} = {2^{32.2}}^{32}$$..how ?

sorry for the silly question, ..cannot understand how $$32^x = 32.x$$ please help , again sorry

$$(a^m)^n=a^{mn}$$, so $$x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}$$.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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14 Oct 2012, 23:54
Bunuel wrote:
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

Given: $$x>0$$ and $$x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}$$ --> $$x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}$$, as the expression under the square root extends infinitely then expression in brackets would equal to $$x$$ itself and we can safely replace it with $$x$$ and rewrite the given expression as $$x=\sqrt{6+x}$$. Square both sides: $$x^2=6+x$$ --> $$(x+2)(x-3)=0$$ --> $$x=-2$$ or $$x=3$$, but since $$x>0$$ then: $$x=3$$.

Hi Bunuel - All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate?

$$x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}$$, as the expression under the square root extends infinitely then expression in brackets would equal to $$x$$ itself and we can safely replace it with $$x$$ and rewrite the given expression as $$x=\sqrt{6+x}$$.

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

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15 Oct 2012, 02:09
Expert's post
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This post was
BOOKMARKED
Jp27 wrote:
Bunuel wrote:
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

Given: $$x>0$$ and $$x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}$$ --> $$x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}$$, as the expression under the square root extends infinitely then expression in brackets would equal to $$x$$ itself and we can safely replace it with $$x$$ and rewrite the given expression as $$x=\sqrt{6+x}$$. Square both sides: $$x^2=6+x$$ --> $$(x+2)(x-3)=0$$ --> $$x=-2$$ or $$x=3$$, but since $$x>0$$ then: $$x=3$$.

Hi Bunuel - All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate?

$$x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}$$, as the expression under the square root extends infinitely then expression in brackets would equal to $$x$$ itself and we can safely replace it with $$x$$ and rewrite the given expression as $$x=\sqrt{6+x}$$.

Cheers

Given: $$x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}$$. Consier the expression in brackets: $${(\sqrt{6+\sqrt{6+\sqrt{6+...}})$$. It's the same as the right hand side of the initial expression, thus it also equals to $$x$$. When replaced we'll have: $$x=\sqrt{6+x}$$.

Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Nov 2012, 05:21
Bunuel wrote:
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 geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first $$n$$ terms of geometric progression is given by: $$sum=\frac{b*(r^{n}-1)}{r-1}$$, where $$b$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$\neq{1}$$.

So in our case: $$5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6$$.

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: $$sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}$$, so the actual sum must be less than 5^7, thus the answer must be A: 5^6.

I am not familiar with GP, could you please let me know how we have consider "r". Using other examples???
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Nov 2012, 05:30
breakit wrote:
Bunuel wrote:
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 geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first $$n$$ terms of geometric progression is given by: $$sum=\frac{b*(r^{n}-1)}{r-1}$$, where $$b$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$\neq{1}$$.

So in our case: $$5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6$$.

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: $$sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}$$, so the actual sum must be less than 5^7, thus the answer must be A: 5^6.

I am not familiar with GP, could you please let me know how we have consider "r". Using other examples???

Check here: sequences-progressions-101891.html
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Nov 2012, 05:55
Bunuel wrote:
breakit wrote:
Bunuel wrote:
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 geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first $$n$$ terms of geometric progression is given by: $$sum=\frac{b*(r^{n}-1)}{r-1}$$, where $$b$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$\neq{1}$$.

So in our case: $$5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6$$.

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: $$sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}$$, so the actual sum must be less than 5^7, thus the answer must be A: 5^6.

I am not familiar with GP, could you please let me know how we have consider "r". Using other examples???

Check here: sequences-progressions-101891.html

Thanks a lot ... I suppose this is not available in Gmatclub math book
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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03 Nov 2012, 05:58
breakit wrote:

Thanks a lot ... I suppose this is not available in Gmatclub math book

Actually it is, check Sequences and Progressions here: gmat-math-book-87417.html
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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13 Jan 2013, 23:38
Thanks Bunuel for these interesting problems.

With respect to problem number 2

"So, we have that the units digit of is 1 and the units digit of is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of is 2."

I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number

eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.

I am unable to think any other way. Please let me know where I am going wrong.

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

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14 Jan 2013, 00:20
gmatrant wrote:
Thanks Bunuel for these interesting problems.

With respect to problem number 2

"So, we have that the units digit of is 1 and the units digit of is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of is 2."

I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number

eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.

I am unable to think any other way. Please let me know where I am going wrong.

Thanks
gmatrant

First of all 757,571 - 58,299,374,483 = -58,298,616,912

You could consider easier cases: 11-13=-2, 11-23=-12, 21-83=-62, ...
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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21 Jan 2013, 09:03
This is as stated correctly a tricky Question. But if we do a little arranging before we start to solve the problem then we can eliminate the confusion. From the question we can easily determine that 1973^9 is greater than 17^12. So lets rewrite the question as -(1973^9 - 17^12). And now no body will be confusing this and the answer is 2 clearly.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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23 Mar 2013, 00:25
2
KUDOS
Bunuel wrote:
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 negative number is 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 --> $$x=\sqrt[5]{-37}\approx{-2.1}<-2$$. Thus $$x^3\approx{(-2.1)^3}\approx{-8.something}<-8$$, so option D must be true.

As for the other options:
A. $$\sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1}<2$$, $$\sqrt{-x}>2$$ is not true.
B. $$x\approx{-2.1}<-2$$, thus x>-2 is also not true.
C. $$x^2\approx{(-2.1)}^2=4.something>4$$, thus x^2<4 is also not true.
E. $$x^4\approx{(-2.1)}^4\approx17$$, (2^4=16, so anyway -2.1^4 can not be more than 32) thus x^4>32 is also not true.

Please correct the GC test Q : M26-05. Error highlighted in red

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

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23 Mar 2013, 04:21
greatps24 wrote:
Bunuel wrote:
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 negative number is 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 --> $$x=\sqrt[5]{-37}\approx{-2.1}<-2$$. Thus $$x^3\approx{(-2.1)^3}\approx{-8.something}<-8$$, so option D must be true.

As for the other options:
A. $$\sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1}<2$$, $$\sqrt{-x}>2$$ is not true.
B. $$x\approx{-2.1}<-2$$, thus x>-2 is also not true.
C. $$x^2\approx{(-2.1)}^2=4.something>4$$, thus x^2<4 is also not true.
E. $$x^4\approx{(-2.1)}^4\approx17$$, (2^4=16, so anyway -2.1^4 can not be more than 32) thus x^4>32 is also not true.

Please correct the GC test Q : M26-05. Error highlighted in red

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

Edited. Thank you.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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30 Mar 2013, 10:53
Great questions and great explanations! Thanks a lot this is really great for practicing. However, with a few of these problems I am running in to the same problem that I have with some of the questions in the gmatclub test's. For example the question below. Even if I assume it only takes me only seconds to read the question 5 seconds to figure out how to handle it: I find it very hard to finish in 2 minutes (litterarly even by copying your answer it almost takes me 2 minutes...) Does this mean I just need to learn to go quicker or are these questions kind of long indeed ? Thanks a lot in advance for your time

Bunuel wrote:
Stiv wrote:
Bunuel wrote:
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: $$\sqrt{60}=2\sqrt{15}$$.

I don't understand why we should un-square the value at the end?

Please read the solution: "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."
Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 30 Mar 2013, 10:53

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