It is currently 17 Dec 2017, 06:04

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 42648

Kudos [?]: 135946 [53], given: 12716

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

### Show Tags

12 Jan 2012, 02:03
53
KUDOS
Expert's post
297
This post was
BOOKMARKED
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
_________________

Kudos [?]: 135946 [53], given: 12716

Manager
Joined: 16 Oct 2012
Posts: 55

Kudos [?]: 8 [0], given: 0

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

### Show Tags

22 Jul 2013, 07:13
Understood, many thanks. Really appreciate it!

Kudos [?]: 8 [0], given: 0

Intern
Status: Finance Analyst
Affiliations: CPA Australia
Joined: 10 Jul 2012
Posts: 18

Kudos [?]: 58 [0], given: 4

Location: Australia
Concentration: Finance, Healthcare
Schools: AGSM '16 (A)
GMAT 1: 470 Q38 V19
GMAT 2: 600 Q44 V34
GPA: 3.5
WE: Accounting (Health Care)
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

22 Jul 2013, 08:28
Edit: 
[/edit]
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.

Thanks for the explanation. The part where you take 5+4 out of the full equation is still unclear to me - 5+4(5+5^2+5^3+5^4+5^5). Could you please break it down into multiple steps?
_________________

Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us.

Your playing small does not serve the world. There's nothing enlightened about shrinking so that other people won't feel insecure around you.

It's not just in some of us; it's in everyone. And as we let our own light shine, we unconsciously give other people permission to do the same.

As we are liberated from our own fear, our presence automatically liberates others.

—Marianne Williamson

Kudos [?]: 58 [0], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 42648

Kudos [?]: 135946 [0], given: 12716

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

### Show Tags

22 Jul 2013, 08:39
vaishnogmat wrote:
Edit: 
[/edit]
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.

Thanks for the explanation. The part where you take 5+4 out of the full equation is still unclear to me - 5+4(5+5^2+5^3+5^4+5^5). Could you please break it down into multiple steps?

4 is factored out from 4*5+4*5^2+4*5^3+4*5^4+4*5^5, not 5+4 from the whole expression: $$4*5+4*5^2+4*5^3+4*5^4+4*5^5=4(5+5^2+5^3+5^4+5^5)$$.

So, $$5+(4*5+4*5^2+4*5^3+4*5^4+4*5^5)=5+4(5+5^2+5^3+5^4+5^5)$$.

Hope it's clear.
_________________

Kudos [?]: 135946 [0], given: 12716

Intern
Joined: 09 May 2013
Posts: 1

Kudos [?]: [0], given: 3

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

### Show Tags

13 Aug 2013, 05:35
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

Hi Bunuel,

It is really a Great post but i think this question can be answered more easily.

When it comes to sq. root of surds $$\sqrt{a +- 2\sqrt{b}}$$ , it can be easily solved if a and b are in the form of a = x +y and b = x*y

For ex. $$\sqrt{25+10\sqrt{6}$$ = $$\sqrt{25+2\sqrt{150}$$ , here 25 = 15 + 10 & 150 = 15*10 so the sqrt will be $$\sqrt{15}$$ + $$\sqrt{10}$$

For this problem , solving similarly , we'll get $$\sqrt{15}$$ + $$\sqrt{10}$$ + $$\sqrt{15}$$ - $$\sqrt{10}$$ = 2$$\sqrt{15}$$

Kudos [?]: [0], given: 3

Intern
Joined: 23 Jul 2013
Posts: 4

Kudos [?]: [0], given: 9

Schools: ISB '15
GMAT 1: 730 Q51 V38
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

14 Aug 2013, 11:05
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.

Hi Bunuel,
I think the answer here has to be 8 instead of 2.

Kudos [?]: [0], given: 9

Math Expert
Joined: 02 Sep 2009
Posts: 42648

Kudos [?]: 135946 [0], given: 12716

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

### Show Tags

14 Aug 2013, 11:11
ashivapu 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.

Hi Bunuel,
I think the answer here has to be 8 instead of 2.

_________________

Kudos [?]: 135946 [0], given: 12716

Manager
Joined: 06 Jun 2012
Posts: 140

Kudos [?]: 284 [0], given: 37

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

### Show Tags

16 Aug 2013, 08:07
Hi,
Need some clarification for the below question.
When 1-3= -2 . So should the value be mod?

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.

_________________

Please give Kudos if you like the post

Kudos [?]: 284 [0], given: 37

Manager
Joined: 20 Jul 2012
Posts: 156

Kudos [?]: 42 [0], given: 559

Location: India
WE: Information Technology (Computer Software)
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

16 Aug 2013, 21:11
1) Did it by approximation...

sqrt6 is almost equal to 2.4
10 sqrt6 is 24
so 25+10sqrt6=49
sqrt{25+10sqrt6}=7

now
25-10sqrt6=1
sqrt{25-10sqrt66}=1

sqrt{25+10sqrt6} + sqrt{25-10sqrt6} is approximately 8
A,D,E can be immediately eliminated...Down to B and C..
and C has a value near to 8

Hope it helps...
_________________

Preparing for another shot...

Kudos [?]: 42 [0], given: 559

Manager
Joined: 20 Jul 2012
Posts: 156

Kudos [?]: 42 [0], given: 559

Location: India
WE: Information Technology (Computer Software)
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

16 Aug 2013, 23:07
ashivapu 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.

Hi Bunuel,
I think the answer here has to be 8 instead of 2.

Hii Bunel
I am clear till the second last line i.e 11-13=-2.. But couldn't conclude (17^3)^4-1973^{3^2}[/m] is 2...Could you please explain
_________________

Preparing for another shot...

Kudos [?]: 42 [0], given: 559

Math Expert
Joined: 02 Sep 2009
Posts: 42648

Kudos [?]: 135946 [1], given: 12716

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

### Show Tags

17 Aug 2013, 02:04
1
KUDOS
Expert's post
akankshasoneja wrote:
ashivapu 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.

Hi Bunuel,
I think the answer here has to be 8 instead of 2.

Hii Bunel
I am clear till the second last line i.e 11-13=-2.. But couldn't conclude (17^3)^4-1973^{3^2}[/m] is 2...Could you please explain

Check here:
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
_________________

Kudos [?]: 135946 [1], given: 12716

Manager
Joined: 20 Jul 2012
Posts: 156

Kudos [?]: 42 [0], given: 559

Location: India
WE: Information Technology (Computer Software)
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

17 Aug 2013, 04:12
Thanks Bunuel...
_________________

Preparing for another shot...

Kudos [?]: 42 [0], given: 559

Intern
Joined: 23 Jul 2013
Posts: 4

Kudos [?]: [0], given: 9

Schools: ISB '15
GMAT 1: 730 Q51 V38
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

21 Aug 2013, 03:44
Bunuel wrote:
ashivapu 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.

Hi Bunuel,
I think the answer here has to be 8 instead of 2.

I had missed this part: 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.

Thanks for the clarification

Kudos [?]: [0], given: 9

Intern
Joined: 28 Jan 2013
Posts: 5

Kudos [?]: 7 [0], given: 100

Location: United States
Concentration: General Management, Entrepreneurship
Schools: HBS '17
GMAT Date: 10-29-2014
GPA: 2.9
WE: Programming (Computer Software)
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

01 Nov 2013, 12:10
If 5 ^10x =4900 and 2sqrt(y)=25 what is the value of (5^(x-1))5/4^-sqrt(y)?

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

5^10x = 4900 which can be written as (5^5x)^2 = 70^2 which implies 5^5x = 70
2^sqrt(y)= 25 or 5^2

Now (5^(x-1))5/4^-sqrt(y) can be written as (5^5x/5^5)/((1/2^y)^2) which is equal to (70/5^5)/(1/5^2)2 = 70/5= 14

Kudos [?]: 7 [0], given: 100

Current Student
Joined: 26 Dec 2011
Posts: 6

Kudos [?]: 3 [0], given: 3

Schools: Yale '17 (A)
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

17 Dec 2013, 18:51
Bunuel wrote:
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

Rearranging both expressions we'll get: $$5x-(z-3y)=125$$ and $$\sqrt{5x}-\sqrt{z-3y}=5$$. Denote $$\sqrt{5x}$$ as $$a$$ and $$\sqrt{z-3y}$$ as $$b$$.

So we have that $$a^2-b^2=125$$ and $$a-b=5$$. Now, $$a^2-b^2=(a-b)(a+b)=125$$ and as $$a-b=5$$ then $$(a-b)(a+b)=5*(a+b)=125$$ --> $$a+b=25$$. Thus we get two equations with two unknowns: $$a+b=25$$ and $$a-b=5$$ --> solving for $$a$$ --> $$a=15=\sqrt{5x}$$ --> $$x=45$$. Solving for $$b$$ -->$$b=10=\sqrt{z-3y}$$

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

This is how I went about solving this problem:

Given $$\sqrt{5x} - 5 - \sqrt{z-3y}= 0$$:

Rearrange and square
$$(\sqrt{5x})^2 = (\sqrt{z-3y}+5)^2$$
$$|5x| = z-3y + 25 + 10\sqrt{z-3y}$$

Assuming that x is positive and given $$5x=125-3y+z$$, set the two equations equal to each other
$$125 - 3y + z = z-3y + 25 + 10\sqrt{z-3y}$$
$$100 = 10\sqrt{z-3y}$$
$$100 = z-3y$$

$$5x = 125-3y+z$$
$$5x = 125 + 100$$
$$x = 45$$

Thus by substitution:
$$= \sqrt{\frac{45(z-3y)}{x}}$$

$$= \sqrt{\frac{45(100)}{45}}$$

$$= \sqrt{100}$$

$$= 10$$

However, notice that I had to assume that |5x| = 5x
Why is this permissible in this problem?

Kudos [?]: 3 [0], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 42648

Kudos [?]: 135946 [1], given: 12716

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

### Show Tags

18 Dec 2013, 01:04
1
KUDOS
Expert's post
NateTheGreat11 wrote:
Bunuel wrote:
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

Rearranging both expressions we'll get: $$5x-(z-3y)=125$$ and $$\sqrt{5x}-\sqrt{z-3y}=5$$. Denote $$\sqrt{5x}$$ as $$a$$ and $$\sqrt{z-3y}$$ as $$b$$.

So we have that $$a^2-b^2=125$$ and $$a-b=5$$. Now, $$a^2-b^2=(a-b)(a+b)=125$$ and as $$a-b=5$$ then $$(a-b)(a+b)=5*(a+b)=125$$ --> $$a+b=25$$. Thus we get two equations with two unknowns: $$a+b=25$$ and $$a-b=5$$ --> solving for $$a$$ --> $$a=15=\sqrt{5x}$$ --> $$x=45$$. Solving for $$b$$ -->$$b=10=\sqrt{z-3y}$$

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

This is how I went about solving this problem:

Given $$\sqrt{5x} - 5 - \sqrt{z-3y}= 0$$:

Rearrange and square
$$(\sqrt{5x})^2 = (\sqrt{z-3y}+5)^2$$
$$|5x| = z-3y + 25 + 10\sqrt{z-3y}$$

Assuming that x is positive and given $$5x=125-3y+z$$, set the two equations equal to each other
$$125 - 3y + z = z-3y + 25 + 10\sqrt{z-3y}$$
$$100 = 10\sqrt{z-3y}$$
$$100 = z-3y$$

$$5x = 125-3y+z$$
$$5x = 125 + 100$$
$$x = 45$$

Thus by substitution:
$$= \sqrt{\frac{45(z-3y)}{x}}$$

$$= \sqrt{\frac{45(100)}{45}}$$

$$= \sqrt{100}$$

$$= 10$$

However, notice that I had to assume that |5x| = 5x
Why is this permissible in this problem?

Even roots from negative numbers are undefined for the GMAT. We have $$\sqrt{5x}$$ in one of the expressions, which means that x is non-negative, thus |5x|=5x.

Hope it's clear.
_________________

Kudos [?]: 135946 [1], given: 12716

Manager
Joined: 30 Mar 2013
Posts: 130

Kudos [?]: 58 [0], given: 196

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

### Show Tags

23 May 2014, 23:46
Bunuel 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. $$2^{22x}*11^{22x}$$

Note that we need approximate value of the given expression. Now, $$22^{22x}$$ is much larger number than $$22^{2x}$$. Hence $$22^{22x}-22^{2x}$$ will be very close to $$22^{22x}$$ itself, basically $$22^{2x}$$ is negligible in this case. The same way $$11^{11x}-11^x$$ will be very close to $$11^{11x}$$ itself.

Thus $$\frac{22^{22x}-22^x}{11^{11x}-11^x}\approx{\frac{22^{22x}}{11^{11x}}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$.

You can check this algebraically as well: $$\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}=\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}$$. Again, -1, both in denominator and nominator is negligible value and we'll get the same expression as above: $$\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\approx{\frac{22^{2x}*22^{20x}}{11^x*11^{10x}}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$

Hello, I don't understand how the division at the end became a multiplication. Could someone help please?

Kudos [?]: 58 [0], given: 196

Intern
Joined: 13 May 2014
Posts: 38

Kudos [?]: 77 [1], given: 1

Concentration: General Management, Strategy
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

### Show Tags

24 May 2014, 00:13
1
KUDOS
usre123 wrote:
Bunuel 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. $$2^{22x}*11^{22x}$$

Note that we need approximate value of the given expression. Now, $$22^{22x}$$ is much larger number than $$22^{2x}$$. Hence $$22^{22x}-22^{2x}$$ will be very close to $$22^{22x}$$ itself, basically $$22^{2x}$$ is negligible in this case. The same way $$11^{11x}-11^x$$ will be very close to $$11^{11x}$$ itself.

Thus $$\frac{22^{22x}-22^x}{11^{11x}-11^x}\approx{\frac{22^{22x}}{11^{11x}}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$.

You can check this algebraically as well: $$\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}=\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}$$. Again, -1, both in denominator and nominator is negligible value and we'll get the same expression as above: $$\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\approx{\frac{22^{2x}*22^{20x}}{11^x*11^{10x}}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$

Hello, I don't understand how the division at the end became a multiplication. Could someone help please?

Hi usre123,

$$\frac{22^{22x}}{11^{11x}} = \frac{(2*11)^{22x}}{11^{11x}} = \frac{(2)^{22x}*(11)^{22x}}{11^{11x}} = 2^{22x}*11^{(22x-11x)} = 2^{22x}*11^{11x}$$

Hope it helps
Press kudos if it helped.

Kindly attempt the question:
if-a-b-1-3-b-c-2-c-d-1-2-d-e-3-and-e-f-1-4-the-171692.html#p1367609

Kudos is the best form of appreciation

Kudos [?]: 77 [1], given: 1

Manager
Joined: 30 Mar 2013
Posts: 130

Kudos [?]: 58 [0], given: 196

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

### Show Tags

24 May 2014, 05:36
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.

Hello...I just dont get it....if its 1-3, wont you borrow from the tens and it will be 11-3..which makes it 8? I'm sorry I dont get it

Kudos [?]: 58 [0], given: 196

Manager
Joined: 30 Mar 2013
Posts: 130

Kudos [?]: 58 [0], given: 196

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

### Show Tags

24 May 2014, 05:38
gmatacequants wrote:
usre123 wrote:
Bunuel 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. $$2^{22x}*11^{22x}$$

Note that we need approximate value of the given expression. Now, $$22^{22x}$$ is much larger number than $$22^{2x}$$. Hence $$22^{22x}-22^{2x}$$ will be very close to $$22^{22x}$$ itself, basically $$22^{2x}$$ is negligible in this case. The same way $$11^{11x}-11^x$$ will be very close to $$11^{11x}$$ itself.

Thus $$\frac{22^{22x}-22^x}{11^{11x}-11^x}\approx{\frac{22^{22x}}{11^{11x}}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$.

You can check this algebraically as well: $$\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}=\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}$$. Again, -1, both in denominator and nominator is negligible value and we'll get the same expression as above: $$\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\approx{\frac{22^{2x}*22^{20x}}{11^x*11^{10x}}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$

Hello, I don't understand how the division at the end became a multiplication. Could someone help please?

Hi usre123,

$$\frac{22^{22x}}{11^{11x}} = \frac{(2*11)^{22x}}{11^{11x}} = \frac{(2)^{22x}*(11)^{22x}}{11^{11x}} = 2^{22x}*11^{(22x-11x)} = 2^{22x}*11^{11x}$$

Hope it helps
Press kudos if it helped.

Kindly attempt the question:
if-a-b-1-3-b-c-2-c-d-1-2-d-e-3-and-e-f-1-4-the-171692.html#p1367609

Kudos is the best form of appreciation

It absolutely did! Please attempt the one below as well! thank you!

Kudos [?]: 58 [0], given: 196

Math Expert
Joined: 02 Sep 2009
Posts: 42648

Kudos [?]: 135946 [0], given: 12716

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

### Show Tags

24 May 2014, 05:41
usre123 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.

Hello...I just dont get it....if its 1-3, wont you borrow from the tens and it will be 11-3..which makes it 8? I'm sorry I dont get it

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
_________________

Kudos [?]: 135946 [0], given: 12716

Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 24 May 2014, 05:41

Go to page   Previous    1   2   3   4   5   6   7   8   9   10    Next  [ 186 posts ]

Display posts from previous: Sort by