GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2019, 15:56 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  NEW!!! Tough and tricky exponents and roots questions

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Manager  Status: Employed
Joined: 17 Nov 2011
Posts: 78
Location: Pakistan
Concentration: International Business, Marketing
GMAT 1: 720 Q49 V40 GPA: 3.2
WE: Business Development (Internet and New Media)
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

2
7
Q3.

$$5^{10x}=4900$$
so $${(5^{5x})}^2=70^2$$

Take square root of both sides >

so $$5^{5x}=70$$

Also from question stem:

$$2^{\sqrt{y}} = 5^2$$

So (from the question stem)

$$4^{-\sqrt{y}} = \frac{1}{4^{\sqrt{y}}}$$
$$4^{-\sqrt{y}} = (\frac{1}{2^{\sqrt{y}}})^2 = \frac{1}{5^4}$$

Now $$5^{(x-1)^5}$$

Should be $$\frac{5^{5x}}{5^5}$$

We know $$5^{10x}=4900$$

so $$5^{10x}=70^2$$

so $$5^{5x}=70$$

So $$\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}$$

is basically $$\frac{5^x}{5^5}*5^4$$

We already know the value of $$5^x$$ which is $$70$$

So now it becomes $$\frac{70}{5^5}*5^4$$

Which should resolve to $$14$$

Hence Answer = E = $$14$$
_________________
"Nowadays, people know the price of everything, and the value of nothing." Oscar Wilde
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

19
48
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

25
46
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.

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

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

19
46
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

27
48
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.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

13
35
6. If $$x=\sqrt{-37}$$ then which of the following must be true?
A. $$\sqrt{-x}>2$$
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32

Must know for the GMAT: Even roots from negative number is undefined on the GMAT (as GMAT is dealing only with Real Numbers): $$\sqrt[{even}]{negative}=undefined$$, for example $$\sqrt{-25}=undefined$$.

Odd roots have the same sign as the base of the root. For example, $$\sqrt{125} =5$$ and $$\sqrt{-64} =-4$$.

Back to the original question:

As $$-2^5=-32$$ then $$x$$ must be a little bit less than -2 --> $$x=\sqrt{-37}\approx{-2.1}<-2$$. Thus $$x^3\approx{(-2.1)^3}\approx{-8.something}<-8$$, so option D must be true.

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

16
1
56
7. If $$x=\sqrt{10}+\sqrt{9}+\sqrt{8}+\sqrt{7}+\sqrt{6}+\sqrt{5}+\sqrt{4}+\sqrt{3}+\sqrt{2}$$, 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

Here is a little trick: any positive integer root from a number more than 1 will be more than 1. For example: $$\sqrt{2}>1$$.

Now, $$\sqrt{10}>3$$ (as 3^2=9) and $$\sqrt{9}>2$$ (2^3=8). Thus $$x=(# \ more \ then \ 3)+(# \ more \ then \ 2)+(7 \ numbers \ more \ then \ 1)=$$
$$=(# \ more \ then \ 5)+(# \ more \ then \ 7)=$$
$$=(# \ more \ then \ 12)$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

19
51
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: $$x=3$$.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

10
44
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: $$\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\approx{\frac{22^{2x}*22^{20x}}{11^x*11^{10x}}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

26
45
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, $$\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10$$.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

17
27
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 $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

_________________
Manager  Joined: 06 Oct 2011
Posts: 162
Schools: Wharton '15, CBS '15
GMAT Date: 06-30-2012
GPA: 3.7
WE: Accounting (Insurance)
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

2
“The only true wisdom is in knowing you know nothing.”
― Socrates

back to studying!
_________________
Reward wisdom with kudos
Manager  Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 79
Concentration: Strategy, General Management
GMAT 1: 460 Q35 V20 GPA: 3.6
WE: Consulting (Computer Software)
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

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?
Please let me know,.
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

2
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?
Please let me know,.

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.
_________________
Intern  Joined: 03 Nov 2010
Posts: 11
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

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...
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

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.
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

kuttingchai wrote:
Bunuel wrote:
2. What is the units digit of $$(17^3)^4-1973^{3^2}$$?
A. 0
B. 2
C. 4
D. 6
E. 8

Must know for the GMAT:
I. The units digit of $$(abc)^n$$ is the same as that of $$c^n$$, which means that the units digit of $$(17^3)^4$$ is that same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is that same as that of $$3^{3^2}$$.

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

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

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

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

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

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

5. 7^5=7 (last digit is 7 again!)
...

1. 3^1=3 (last digit is 3)
2. 3^2=9 (last digit is 9)
3. 3^3=27 (last digit is 7)
4. 3^4=81 (last digit is 1)

5. 3^5=243 (last digit is 3 again!)
...

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

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

Hey Bunuel,

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

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

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

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

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

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

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

Hope i's clear.
_________________
Director  Joined: 22 Mar 2011
Posts: 588
WE: Science (Education)
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

Bunuel wrote:
SOLUTIONS:

1. What is the value of $$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}$$?
A. $$2\sqrt{5}$$
B. $$\sqrt{55}$$
C. $$2\sqrt{15}$$
D. 50
E. 60

Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know fro the GMAT: $$(x+y)^2=x^2+2xy+y^2$$ (while $$(x-y)^2=x^2-2xy+y^2$$).

So we get: $$(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=$$
$$=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6})$$.

Note that sum of the first and the third terms simplifies to $$(25+10\sqrt{6})+(25-10\sqrt{6})=50$$, so we have $$50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})$$ --> $$50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}$$.

Also must know for the GMAT: $$(x+y)(x-y)=x^2-y^2$$, thus $$50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)}=50+2\sqrt{625-600}=50+2\sqrt{25}=60$$.

Recall that we should un-square this value to get the right the answer: $$\sqrt{60}=2\sqrt{15}$$.

Another way to do it, using the same formulas:

$$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{5(5+2\sqrt{6})}+\sqrt{5(5-2\sqrt{6})}=\sqrt{5}\sqrt{5+2\sqrt{6}}+\sqrt{5}\sqrt{5-2\sqrt{6}}=$$
$$=\sqrt{5}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=\sqrt{5}(\sqrt{(\sqrt{3}+\sqrt{2})^2}+\sqrt{(\sqrt{3}-\sqrt{2})^2})=\sqrt{5}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=\sqrt{5}(2\sqrt{3})=2\sqrt{15}$$
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.

Originally posted by EvaJager on 29 Jun 2012, 11:34.
Last edited by EvaJager on 15 Oct 2012, 01:35, edited 1 time in total.
Manager  Joined: 22 Dec 2011
Posts: 211
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

Bunuel wrote:
8. If $$x$$ is a positive number and equals to $$\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}$$, where the given expression extends to an infinite number of roots, then what is the value of x?
A. $$\sqrt{6}$$
B. 3
C. $$1+\sqrt{6}$$
D. $$2\sqrt{3}$$
E. 6

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

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

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

Cheers
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

Show Tags

1
Jp27 wrote:
Bunuel wrote:
8. If $$x$$ is a positive number and equals to $$\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}$$, where the given expression extends to an infinite number of roots, then what is the value of x?
A. $$\sqrt{6}$$
B. 3
C. $$1+\sqrt{6}$$
D. $$2\sqrt{3}$$
E. 6

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

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

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

Cheers

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

Hope it's clear.
_________________ Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 15 Oct 2012, 03:09

Go to page   Previous    1   2   3   4   5    Next  [ 98 posts ]

Display posts from previous: Sort by

NEW!!! Tough and tricky exponents and roots questions

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  