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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

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

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

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

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. \(2^{22x}*11^{22x}\)

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

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

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24 May 2014, 07:10

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Bunuel wrote:

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.

Answer B.

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

Thank you, but I do have one more question, if anyone could please explain. HOW do I know first part is smaller than second....I understand why its not 8 (cause first digit, 17^3^4 is smaller than the answer for the second part, so we do 1-3 and take the absolute). And we do 11-3 (borrow from tens) when first part is larger than the second....so 11-3. But when do I know which is which? I can t possible tell which digit is larger from looking at the question.

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.

Answer B.

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

Thank you, but I do have one more question, if anyone could please explain. HOW do I know first part is smaller than second....I understand why its not 8 (cause first digit, 17^3^4 is smaller than the answer for the second part, so we do 1-3 and take the absolute). And we do 11-3 (borrow from tens) when first part is larger than the second....so 11-3. But when do I know which is which? I can t possible tell which digit is larger from looking at the question.

This question is addressed in the first link given above:

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 \(1000^{(3^2)}=1,000^9=10^{27}\) --> \(10^{24}<10^{27}\).

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

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11 Jun 2014, 01:46

5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2

How come you arrive to 25 from y?

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.

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.

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

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11 Jun 2014, 02:09

\((2^{\sqrt{y}})^2=25^2\)

Bunuel explain this part please

Bunuel wrote:

arindamsur wrote:

5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2

How come you arrive to 25 from y?

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.

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.

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

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11 Jun 2014, 13:55

Can someone please show me the breakdown of this "4(5(5^5-1)/5-1)=5^6"

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.

Can someone please show me the breakdown of this "4(5(5^5-1)/5-1)=5^6"

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.

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

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11 Jun 2014, 14:08

Yes; thank you very much.

Bunuel wrote:

farhanabad wrote:

Can someone please show me the breakdown of this "4(5(5^5-1)/5-1)=5^6"

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.

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20 Aug 2014, 12:53

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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.

Answer: A.

other solution: The sum = \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5 =\)

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

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03 Oct 2014, 22:23

Hi Bunuel, I Have a question(might be silly) here given is \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}

i do not understand how the above equation breaking into below-

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}).

Thanks for your help.

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: \(\sqrt{60}=2\sqrt{15}\).

Answer: C.

_________________

--------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way

Hi Bunuel, I Have a question(might be silly) here given is \sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}

i do not understand how the above equation breaking into below-

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}).

Thanks for your help.

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: \(\sqrt{60}=2\sqrt{15}\).

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.

Answer: C.

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.

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.

Answer: C.

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.

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?

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.

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

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11 Jul 2015, 16:28

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This post was BOOKMARKED

Bunuel wrote:

ranaazad wrote:

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.

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

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13 Sep 2015, 21:18

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.

Answer: A.

how can i understand that all terms are equal to the largest term