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

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Math Expert
Joined: 02 Sep 2009
Posts: 44657
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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Joined: 27 Oct 2014
Posts: 8
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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07 Dec 2014, 11:20
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Bunuel wrote:
5. If $$x=23^2*25^4*27^6*29^8$$ and is a multiple of $$26^n$$, where $$n$$ is a non-negative integer, then what is the value of $$n^{26}-26^n$$?
A. -26
B. -25
C. -1
D. 0
E. 1

$$23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd$$ so $$x$$ is an odd number. The only way it to be a multiple of $$26^n$$ (even number in integer power) is when $$n=0$$, in this case $$26^n=26^0=1$$ and 1 is a factor of every integer. Thus $$n=0$$ --> $$n^{26}-26^n=0^{26}-26^0=0-1=-1$$. Must know for the GMAT: $$a^0=1$$, for $$a\neq{0}$$ - any nonzero number to the power of 0 is 1. Important note: the case of 0^0 is not tested on the GMAT.

If 13 is the biggest prime factor of 26 and there isn't a 13 prime factor in X, then 26^0 is the only answer.

Is this approach correct?

Many thanks,
Math Expert
Joined: 02 Sep 2009
Posts: 44657
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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08 Dec 2014, 03:10
joaogallegomoura wrote:
Bunuel wrote:
5. If $$x=23^2*25^4*27^6*29^8$$ and is a multiple of $$26^n$$, where $$n$$ is a non-negative integer, then what is the value of $$n^{26}-26^n$$?
A. -26
B. -25
C. -1
D. 0
E. 1

$$23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd$$ so $$x$$ is an odd number. The only way it to be a multiple of $$26^n$$ (even number in integer power) is when $$n=0$$, in this case $$26^n=26^0=1$$ and 1 is a factor of every integer. Thus $$n=0$$ --> $$n^{26}-26^n=0^{26}-26^0=0-1=-1$$. Must know for the GMAT: $$a^0=1$$, for $$a\neq{0}$$ - any nonzero number to the power of 0 is 1. Important note: the case of 0^0 is not tested on the GMAT.

If 13 is the biggest prime factor of 26 and there isn't a 13 prime factor in X, then 26^0 is the only answer.

Is this approach correct?

Many thanks,

______________
Yes, that's correct.
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Joined: 16 Jan 2013
Posts: 95
GMAT 1: 490 Q41 V18
GMAT 2: 610 Q45 V28
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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10 Jul 2015, 17:55
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.

Hello Bunuel,
Nice explanations. I am a little confused with the fact which you mentioned as "negligible" in this math. Wouldn't this deduction change the result of the answer?

Thanks.
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Math Expert
Joined: 02 Sep 2009
Posts: 44657
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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11 Jul 2015, 02:40
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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.

Hello Bunuel,
Nice explanations. I am a little confused with the fact which you mentioned as "negligible" in this math. Wouldn't this deduction change the result of the answer?

Thanks.

The question asks about the closest value of the fraction among the options, not the exact value, so we can approximate.

Similar questions to practice:
the-value-of-10-8-10-2-10-7-10-3-is-closest-to-which-of-95082.html
which-of-the-following-is-closest-to-10180-1030-a-110224.html
which-of-the-following-best-approximates-the-value-of-q-if-99674.html
new-tough-and-tricky-exponents-and-roots-questions-125956-40.html
if-x-10-10-x-2-2x-7-3x-2-10x-2-is-closest-to-143897.html
1-0-0001-0-04-10-the-value-of-the-expression-above-is-59398.html
which-of-the-following-is-closest-in-value-to-64425.html
10-180-10-30-which-of-the-following-best-approximates-84309.html
if-x-2-b-8-8-8-6-for-which-of-the-following-b-values-160197.html
if-x-3-000-then-the-value-of-x-2x-1-is-closest-to-166128.html

Hope it helps.
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Joined: 04 Jun 2016
Posts: 617
GMAT 1: 750 Q49 V43
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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Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
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Posts: 10
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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14 Feb 2018, 09:40
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.

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

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

Hey there, just one remark. Why could 4sqrt/y not become by 2 x 2 sqrt/y? why is it definitely (2 sqrt/y)^2 ?

thanks
Math Expert
Joined: 02 Sep 2009
Posts: 44657
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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14 Feb 2018, 09:47
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gmatmo wrote:
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.

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

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

Hey there, just one remark. Why could 4sqrt/y not become by 2 x 2 sqrt/y? why is it definitely (2 sqrt/y)^2 ?

thanks

You can write this in several ways:

$$4^{\sqrt{y}}=(2*2)^{\sqrt{y}}=2^{\sqrt{y}}*2^{\sqrt{y}}=2^{\sqrt{y}+\sqrt{y}}=2^{2\sqrt{y}}=(2^{\sqrt{y}})^2$$

P.S. Please read the following post. Writing Mathematical Formulas on the Forum: https://gmatclub.com/forum/rules-for-po ... l#p1096628
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NEW!!! Tough and tricky exponents and roots questions [#permalink]

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08 Apr 2018, 10:57
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 have a doubt here.
If we are doing this subtraction (1-3). And, if we would have actually solved the whole subtraction of the number. the we would have borrowed 10 from the tens digit. So, the final subtraction would have looked like (11-3)= 8
please let me know if I am correct or not. if not, then why?
Math Expert
Joined: 02 Sep 2009
Posts: 44657
Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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08 Apr 2018, 11:35
1
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Expert's post
shivangibh 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 have a doubt here.
If we are doing this subtraction (1-3). And, if we would have actually solved the whole subtraction of the number. the we would have borrowed 10 from the tens digit. So, the final subtraction would have looked like (11-3)= 8
please let me know if I am correct or not. if not, then why?

You could very easily test cases to prove your logic.

(positive number ending with 1) - (greater number ending with 3) = (negative number ending with 2)

11-23=-12
31-133=-102
41-123=-82
....

I tried to explain this several times in this thread:
https://gmatclub.com/forum/new-tough-an ... l#p1054715
https://gmatclub.com/forum/new-tough-an ... l#p1099223

In addition here is an actual result:

$$(17^3)^4-1973^{3^2}=-453047530293560259230589943252$$
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NEW!!! Tough and tricky exponents and roots questions [#permalink]

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12 Apr 2018, 23:51
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

ALTERNATIVE SOLUTION.
Let $$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}$$=x.
Then $$x^2$$=($$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2$$
And $$x^2$$=$$25+10\sqrt{6}+2\sqrt{625-600}+25-\sqrt{6}$$
$$x^2=60$$ => x=$$2\sqrt{15}$$
Hence C.
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Tulkin.

NEW!!! Tough and tricky exponents and roots questions   [#permalink] 12 Apr 2018, 23:51

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