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NEW!!! Tough and tricky exponents and roots questions

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

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13 Jan 2012, 20:46
3
3
Q4. What is the value of $$5+4*5+4*5^2+4*5^3+4*5^4+4*5^5$$

This one was actually the simplest I thought. Here is how:

$$5+4*5+4*5^2+4*5^3+4*5^4+4*5^5$$

So $$5*(1+4)+4*5^2+4*5^3+4*5^4+4*5^5$$
So $$5^2+4*5^2+4*5^3+4*5^4+4*5^5$$
So $$5^2*(1+4)+4*5^3+4*5^4+4*5^5$$
So $$5^3+4*5^3+4*5^4+4*5^5$$
So $$5^3*(1+4)+4*5^4+4*5^5$$

So every expression behind contributes a power of 1 to the one in front of it. We just need to see the last which is $$5^5$$
Keep solving and you come to a total of $$5^6$$

Hence Answer = $$5^6$$
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14 Jan 2012, 13:53
23
35
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: $$\sqrt{60}=2\sqrt{15}$$.

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14 Jan 2012, 13:58
16
37
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.

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14 Jan 2012, 14:00
23
36
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.

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

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

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14 Jan 2012, 14:03
15
34
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.

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14 Jan 2012, 14:06
12
24
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

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, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

Back to the original question:

As $$-2^5=-32$$ then $$x$$ must be a little bit less than -2 --> $$x=\sqrt[5]{-37}\approx{-2.1}<-2$$. 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.

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14 Jan 2012, 14:08
14
1
44
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

Here is a little trick: any positive integer root from a number more than 1 will be more than 1. For example: $$\sqrt[1000]{2}>1$$.

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

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14 Jan 2012, 14:09
18
42
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: $$x=3$$.

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14 Jan 2012, 14:11
10
35
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: $$\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\approx{\frac{22^{2x}*22^{20x}}{11^x*11^{10x}}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$

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14 Jan 2012, 14:13
25
35
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, $$\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10$$.

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14 Jan 2012, 14:15
17
21
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 $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

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15 Jan 2012, 20:04
2
“The only true wisdom is in knowing you know nothing.”
― Socrates

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

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06 Mar 2012, 22:53
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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07 Mar 2012, 00:36
1
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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23 Apr 2012, 10:50
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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23 Apr 2012, 12:21
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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25 Jun 2012, 08:59
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.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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Updated on: 15 Oct 2012, 00:35
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

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:

$$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{5(5+2\sqrt{6})}+\sqrt{5(5-2\sqrt{6})}=\sqrt{5}\sqrt{5+2\sqrt{6}}+\sqrt{5}\sqrt{5-2\sqrt{6}}=$$
$$=\sqrt{5}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=\sqrt{5}(\sqrt{(\sqrt{3}+\sqrt{2})^2}+\sqrt{(\sqrt{3}-\sqrt{2})^2})=\sqrt{5}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=\sqrt{5}(2\sqrt{3})=2\sqrt{15}$$
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Originally posted by EvaJager on 29 Jun 2012, 10:34.
Last edited by EvaJager on 15 Oct 2012, 00:35, edited 1 time in total.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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14 Oct 2012, 23:54
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
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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15 Oct 2012, 02:09
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.
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Re: NEW!!! Tough and tricky exponents and roots questions &nbs [#permalink] 15 Oct 2012, 02:09

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