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

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06 Mar 2012, 23:53
Quote:
What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8

Quote:
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 note that the second number is mush 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.

I agree with all the steps above but i have this doubt in this

unit digit of 1st number is 1.
Unit digit of second number is 3

1-3
would it not mean
unit digit is 8?
for eg if we have
551 -853 ..Unit digit will be 2
but what if it is 551-453 .unit digit will be 8?

so how do we know which is the larger number of these, although it says X-Y?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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07 Mar 2012, 01:36
1
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Expert's post
shankar245 wrote:
Quote:
What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8

Quote:
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.

I agree with all the steps above but i have this doubt in this

unit digit of 1st number is 1.
Unit digit of second number is 3

1-3
would it not mean
unit digit is 8?
for eg if we have
551 -853 ..Unit digit will be 2
but what if it is 551-453 .unit digit will be 8?

so how do we know which is the larger number of these, although it says X-Y?

That's a little trap there.

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.

So, as you can see the cases like 551-453 (larger number minus smaller number) are not possible.

Why is second number much larger then the first one? Consider this, even if we had (100^3)^4 (instead of (17^3)^4) and 1000^(3^2) (instead of 1973^(3^2)) --> (100^3)^4=100^12=10^24 and 1000^(3^2)=1,000^9=10^27.

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

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07 Mar 2012, 01:57
Thanks..

got it clarified .

I'm finding this really useful as its making me think of short cuts or other ways to solve it..

Nice one Bunuel.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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09 Apr 2012, 13:24
1. c

=========================

2. B

17^12 - 1973 ^ 6

units place for 17^12 = 1
units place for 1973^6 = 9

========================

3. E

5^10x = 4900

(5^5x)^2 = 70^2
5^5x = 70 -------------I

= (5^(x-1))^5 * 4 ^ sqrt y
= 5 ^ (5x-5) * (2 ^ sqrt y) ^ 2
= 5 ^ 5x * 5 ^ -5 * 25 ^2
= (5 ^ 5x / 5 ^ 5) * 5 ^ 4
= 70 / 5 = 14

=========================

4. A

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

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23 Apr 2012, 11:50
Bunuel wrote:
4^{\sqrt{y}

For Question #3.
I'm confused why 4 squareroot Y = (2 squareroot Y)^2 instead of (2 ^2 squareroot Y).

Not sure I understand the properties correctly...
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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23 Apr 2012, 13:21
KGG88 wrote:
Bunuel wrote:
4^{\sqrt{y}

For Question #3.
I'm confused why 4 squareroot Y = (2 squareroot Y)^2 instead of (2 ^2 squareroot Y).

Not sure I understand the properties correctly...

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

Check this for properties: tough-and-tricky-exponents-and-roots-questions-125956.html#p1027888

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

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24 Apr 2012, 12:19
LalaB 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*29^8/26^n= 23^2*25^4*27^6*29^8/(2*13)^n from this point it is obvious, that n =0 (since both 2 and 13 are primes and none of the numbers of numerator can be divisible by 2 and 13)

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

answ is C

is it possible to explain how and why n = 0
had it been 4*12 instead of 2*13, what difference does it make. still the numerator is not divisible. so can we say n=0?
i mean what difference when we have two prime numbers?
i am not getting how n=0 is derived?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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24 Apr 2012, 12:22
kashishh wrote:
LalaB 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*29^8/26^n= 23^2*25^4*27^6*29^8/(2*13)^n from this point it is obvious, that n =0 (since both 2 and 13 are primes and none of the numbers of numerator can be divisible by 2 and 13)

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

answ is C

is it possible to explain how and why n = 0
had it been 4*12 instead of 2*13, what difference does it make. still the numerator is not divisible. so can we say n=0?
i mean what difference when we have two prime numbers?
i am not getting how n=0 is derived?

Below is OE for this question.

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.

Hope it's clear.

P.S. Links to the solutions are given in the initial post.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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24 Apr 2012, 13:17
Bunuel wrote:
kashishh wrote:
LalaB 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*29^8/26^n= 23^2*25^4*27^6*29^8/(2*13)^n from this point it is obvious, that n =0 (since both 2 and 13 are primes and none of the numbers of numerator can be divisible by 2 and 13)

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

answ is C

is it possible to explain how and why n = 0
had it been 4*12 instead of 2*13, what difference does it make. still the numerator is not divisible. so can we say n=0?
i mean what difference when we have two prime numbers?
i am not getting how n=0 is derived?

Below is OE for this question.

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.

Hope it's clear.

P.S. Links to the solutions are given in the initial post.

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

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24 Apr 2012, 14:08
is the answer to the question no. 11 is E?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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24 Apr 2012, 14:11
kashishh wrote:
is the answer to the question no. 11 is E?

Answer to the question # 11 is D.

Again, the links to the solutions and OA's are given in the initial post, here: tough-and-tricky-exponents-and-roots-questions-125956.html#p1027886
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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06 May 2012, 06:12
[quote="Bunuel"]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

Hello Bunnel,
what will be your strategy to do the above question?
Mine was using (a+b)^2=a^2+b^2+2ab but it will some time and calculation is tedious.

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

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06 May 2012, 06:14
balas wrote:
Bunuel wrote:
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

Hello Bunnel,
what will be your strategy to do the above question?
Mine was using (a+b)^2=a^2+b^2+2ab but it will some time and calculation is tedious.

Thanks

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

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

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28 May 2012, 09:37
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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28 May 2012, 09:55
sarikaburnwal wrote:

Links to OA's and solutions are given in the initial post: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1027886

Question #11 is discussed here: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029232

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

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22 Jun 2012, 23:49
Bunuel wrote:
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.

I have "1" and "3" unit digits for first and second statements but 1-3 is like 11-3 = 8 so seems E is correct Answer? Could you explain more?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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23 Jun 2012, 00:43
1
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Bunuel wrote:
but at the same time 5x=125-3y+z 5*45-125=z-3y 10=z-3y
it seems there is a typo in a q stem -\sqrt{5x}-25-\sqrt{z-3y}=0

it must be \sqrt{5x}-5-\sqrt{z-3y}=0

.[/quote]
5*45-125=z-3y ==> 100=z-3y NOT 10 this gives the final answer 10 or B.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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23 Jun 2012, 03:46
omidsa wrote:
Bunuel wrote:
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.

I have "1" and "3" unit digits for first and second statements but 1-3 is like 11-3 = 8 so seems E is correct Answer? Could you explain more?

Please pay attention to the underlined part: 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.

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

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23 Jun 2012, 03:52
omidsa wrote:
but at the same time 5x=125-3y+z 5*45-125=z-3y 10=z-3y
it seems there is a typo in a q stem -\sqrt{5x}-25-\sqrt{z-3y}=0

it must be \sqrt{5x}-5-\sqrt{z-3y}=0

5*45-125=z-3y ==> 100=z-3y NOT 10 this gives the final answer 10 or B
.

I'm not sure I understand what you've written there, so below is a correct solution of the question:

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

Also, please note that the links to OA's and solutions are given in the initial post: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1027886 (if the links does not work, then switch view mode of the topic from "Best Reply" to "Oldest" and then they will lead you to the solutions).
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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25 Jun 2012, 09:15
1
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BOOKMARKED
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??
Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 25 Jun 2012, 09:15

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