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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, 03:03
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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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NEW!!! Tough and tricky exponents and roots questions [#permalink]

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14 Sep 2015, 02:51
anik19890 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.

how can i understand that all terms are equal to the largest term

Look at it via a simple example: let's say you need to calculate the value of 1+2 such that the options are spread wide.

1+1<1+2< 2+2 , thus if there is an option between 2 and 4, that will be your answer.

Similarly for the question above, the sum of the series will be < $$4*5+4*5+4*5^2+4*5^3+4*5^4+4*5^5$$ < $$4*5^6+4*5^6+4*5^6+4*5^6+4*5^6+4*5^6$$ (=$$6*4*5^5=24*5^5$$) < $$25*5^5 (=5^7)$$

Thus the correct answer should be less than $$5^7$$ . Thus A is the correct answer.

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

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04 Dec 2015, 13:31
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 $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

If x^2 = 2^64, then can we write
=> x^2 = $$2^{8^{2}}$$?
And then
=> x = 2^8 ? Since, $$2^{2^{37}}=2^y$$ --> $$y=2^{37}$$.

I know my above solution is not correct, but I don't find where is the mistake.

Pls, help me.

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Last edited by Mahmud6 on 05 Dec 2015, 02:06, edited 1 time in total.

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

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04 Dec 2015, 13:48
Mahmud6 wrote:

x^2 = 2^64
=> x^2 = 2^8^2 (In stack symbol)

=> x = 2^8
I know my above solution is not correct, but I don't find where is the mistake.
Pls, help me.

The mistake you are making is marked in red above.

When you have 2^2 = 2*2 , and you take the square root of this , you get 2 and not $$2^{\sqrt{2}}$$

The reason is $$(2^a)^b = 2^{a*b}$$ and NOT $$2^{a^b}.$$

This is where you are making a mistake.

For the question given,

$$x^2 = 2^{64}$$ ---> $$x = \sqrt{2^{64}} = 2^{64*0.5}$$ (as $$\sqrt{x} = x^{0.5}$$)

Thus $$x = 2^{32}$$

now if $$x^x = 2^y$$ ---> $$(2^{32})^{2^{32}} = 2^y$$

---> From $$(2^a)^b = 2^{a*b}$$ mentioned above,

$$(2^{32})^{2^{32}}= 2^{32*2^{32}} = 2^{2^5*2^{32}} = 2^{2^{37}}$$ (as $$2^a*2^b = 2^{a+b}$$)

Thus, $$2^{2^{37}} = 2^y$$---> y = $$2^{37}$$.

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

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05 Dec 2015, 02:11
Engr2012 wrote:
Mahmud6 wrote:

x^2 = 2^64
=> x^2 = 2^8^2 (In stack symbol)

=> x = 2^8
I know my above solution is not correct, but I don't find where is the mistake.
Pls, help me.

The mistake you are making is marked in red above.

When you have 2^2 = 2*2 , and you take the square root of this , you get 2 and not $$2^{\sqrt{2}}$$

The reason is $$(2^a)^b = 2^{a*b}$$ and NOT $$2^{a^b}.$$

This is where you are making a mistake.

For the question given,

$$x^2 = 2^{64}$$ ---> $$x = \sqrt{2^{64}} = 2^{64*0.5}$$ (as $$\sqrt{x} = x^{0.5}$$)

Thus $$x = 2^{32}$$

now if $$x^x = 2^y$$ ---> $$(2^{32})^{2^{32}} = 2^y$$

---> From $$(2^a)^b = 2^{a*b}$$ mentioned above,

$$(2^{32})^{2^{32}}= 2^{32*2^{32}} = 2^{2^5*2^{32}} = 2^{2^{37}}$$ (as $$2^a*2^b = 2^{a+b}$$)

Thus, $$2^{2^{37}} = 2^y$$---> y = $$2^{37}$$.

Dear Engr2012,

I have edited my problem. Would you pls clarify me?

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

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05 Dec 2015, 08:15
Mahmud6 wrote:

If x^2 = 2^64, then can we write
=> x^2 = $$2^{8^{2}}$$?
And then
=> x = 2^8 ? Since, $$2^{2^{37}}=2^y$$ --> $$y=2^{37}$$.

I know my above solution is not correct, but I don't find where is the mistake.

Pls, help me.

You are making the mistake of separating $$2^8$$ from $$2^{8^2}$$. You can not do that.

You can do $$x^2 = 2^{64}$$ ---->$$x = \sqrt{2^{64}} = 2^{64*0.5}= 2^{32}$$

Also, remember that by definition, $$x^2 = x*x$$ ---> $$(2^8)^2 = 2^8*2^8 = 2^{16}$$ but $$(2^{32})^2= 2^{32}*2^{32} = 2^{64}$$. Thus you see that you will get $$2^{64}$$as a 'squared' entity only when you square $$2^{32}$$.

Coming back to your question, if $$x^2 = 2^{8^2}$$, you must take the square root with the base 2 and NOT for the base 8.

Hope this helps.

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

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15 Dec 2015, 10:55
Dear Bunuel

Thanks for this awesome post.
However I have below doubt:

Sum no:8
Given: x>0 and x=6+6+6+6+...−−−−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√ --> x=6+(6+6+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=6+x−−−−−√.

How is 6+x derived?

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

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15 Dec 2015, 12:23
seemachandran wrote:
Dear Bunuel

Thanks for this awesome post.
However I have below doubt:

Sum no:8
Given: x>0 and x=6+6+6+6+...−−−−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√ --> x=6+(6+6+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=6+x−−−−−√.

How is 6+x derived?

You can look at it this way.

$$x = \sqrt{6+\sqrt{6..}}$$

Square both sides, you get ,

$$x^2= 6+\sqrt{6+\sqrt{6..}}$$ ---> $$x^2-6=x$$ --> $$x^2-x-6=0$$--> $$x=$$3 or $$x= -2$$ .

As x = a square root and for GMAT quant, any quantity with a square root is always >0. Thus neglect x=-2 and hence x=3 is the answer.

Hope this helps.

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

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16 Dec 2015, 20:15
Q7 becomes a bit easier since, options are only x>12.

Would have been tougher if they would had been x>12, 12<x<14,x>14 etc

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16 Dec 2015, 20:16
The most difficult is the first one,

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16 Dec 2015, 20:18
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: $$x=3$$.

Typical question.
Logic:
infinity - 1 = infinity.

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

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21 Mar 2016, 18:45
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.

Since its 1-3. When subtracting the numbers, we will borrow 1 from the ten's digit. Hence it should be 11-3=8. Please help. I believe am going wrong somewhere.

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

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21 Mar 2016, 23:14
atirajak 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.

Since its 1-3. When subtracting the numbers, we will borrow 1 from the ten's digit. Hence it should be 11-3=8. Please help. I believe am going wrong somewhere.

new-tough-and-tricky-exponents-and-roots-questions-125956-60.html#p1054715
new-tough-and-tricky-exponents-and-roots-questions-125956-60.html#p1098765
new-tough-and-tricky-exponents-and-roots-questions-125956-80.html#p1099223
new-tough-and-tricky-exponents-and-roots-questions-125956-60.html#p1054715
new-tough-and-tricky-exponents-and-roots-questions-125956-80.html#p1168559
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NEW!!! Tough and tricky exponents and roots questions [#permalink]

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29 Jun 2016, 11:07
OMG .. there is a much elegant and easier method to solve this

$$\sqrt{6} = ~2.4 ==> 10\sqrt{6}=10*2.4==> 24$$

25+24=49

$$\sqrt{49} =7$$

25-24=1

$$\sqrt{1} = 1$$

So our question is 7+1 = 8

See option $$2 \sqrt{15} = 2*4= 8$$

HENCE C

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: $$\sqrt{60}=2\sqrt{15}$$.

OMG .. there is a much elegant and easier method to solve this

$$\sqrt{6} = ~2.4 ==> 10\sqrt{6}=10*2.4==> 24$$
25+24=49
$$\sqrt{49} =7$$
25-24=1
$$\sqrt{1} = 1$$

So our question is 7+1 = 8

See option $$2 \sqrt{15} = 2*4= 8$$

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

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15 Aug 2016, 10:10
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.

Bunuel ,

I am somewhat unclear. What if you subtract a larger number that has the units digit smaller than the units digit of the smaller number? For example, if you subtract 485 - 28, the answer is 457 which has the unts digit as 7 and not 8-5=3. Can you please help clarify this for me? Thanks

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

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28 Apr 2017, 10:11
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: $$x=3$$.

Hello Bunnel,

Dont understand how you put sqrt(6+x). Not sure of this concept. Any insight would help.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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12 May 2017, 15:55
Solution: http://gmatclub.com/forum/tough-and-tri ... l#p1029232[/quote]

There's a pattern in number 4 called a geometric progression-

$$5+4*5+4*5^2+4*5^3+4*5^4+4*5^5$$?[/b]

It is important to understand the order of operations in this problem first- that is they key to understanding the pattern- multiplication before addition (PEMDAS isn't entirely accurate because both multiplication/division and subtraction/addition are of equal rank therefore whichever appears first is performed first though still both multiplication or division is performed before addition/subtraction

5 + 4(5) =
5 + 20 = 25
25= 5^2

5^2 + 4(5)^2
5^2 + 4(25)
5^2 + 100 = 125
5 ^ 3 = 125

5^3 + (4) 5^3 (remember there is an assumed 1 in front of the 5^3- see adding exponents)
(1) 5^3 + (4) 5^3
( 1 +4) (5^3) - at this point the pattern should become much clearer
(5)(5^3)
5^4

5^4 + (4) 5^4
(1) 5 ^ 4 + (4) 5^4
(1 + 4) 5^4 = 5^5

5^5 + (4) 5^5
(1) 5^5 + (4) 5^5
(1 + 4) 5^5
(5) 5^5

Hence

5^6

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

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12 May 2017, 16:17
8. If xx is a positive number and equals to 6+6+6+6+...−−−−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√6+6+6+6+..., where the given expression extends to an infinite number of roots, then what is the value of x?
A. 6√6
B. 3
C. 1+6√1+6
D. 23√23
E. 6

This problem is a nested radical; more specifically, this problem is a monotonically increasing square root- all this means is that we have to turn the equation into a quadratic formula

6+6+6+6+...−−−−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√6+6+6+6+ = x^2
6+6+6+6+...−−−−−√−−−−−−−−−√−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√6+6+6+6+ = 6 + \sqrt{6}
6 + x = x^2
x^2 + x - 6 = 0

(x - 2) (x+3)

X must be positive

Hence

"B"

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

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08 Jul 2017, 13:55
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: $$x=3$$.

Could someone, please, elaborate on why the infinitely extending expression in brackets equals x itself? This doesn't look trivial to me. Thank you.

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

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25 Sep 2017, 12:32
Bunuel wrote:
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

First thing one should notice here is that $$x$$ and $$y$$ must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

$$5^{10x}=4,900$$ --> $$(5^{5x})^2=70^2$$ --> $$5^{5x}=70$$

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

hi man

The radicals you used in this question is too uncommon for me to read. Please see few things as under:

(1)
how to read this "2y√" (copied from your text) ? Is it "2 to the power root y" ..?
(2)
how to read this "4−y"(copied from your text) ..? Is it 4 to the power minus root y ..?

Most importantly, "x" and "y" must be irrationals. Please shed some light on this concept

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

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26 Sep 2017, 13:40
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 man

Does this expression mean as such as under:

6 to the power 1/2 + (6 to the power 1/4 + 6 to the power 1/8 + 6 to the power 1/16 and so on ....)
and how the expression inside the parenthes equals x?

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Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 26 Sep 2017, 13:40

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