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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 \(\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}}\)?A. \(2\sqrt{5}\) B. \(\sqrt{55}\) C. \(2\sqrt{15}\) D. 50 E. 60 Solution: toughandtrickyexponentsandrootsquestions12595640.html#p10292162. What is the units digit of \((17^3)^41973^{3^2}\)?A. 0 B. 2 C. 4 D. 6 E. 8 Solution: toughandtrickyexponentsandrootsquestions12595640.html#p10292193. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}\)?A. 14/5 B. 5 C. 28/5 D. 13 E. 14 Solution: toughandtrickyexponentsandrootsquestions12595640.html#p10292214. 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: toughandtrickyexponentsandrootsquestions12595640.html#p10292225. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a nonnegative integer, then what is the value of \(n^{26}26^n\)?A. 26 B. 25 C. 1 D. 0 E. 1 Solution: toughandtrickyexponentsandrootsquestions12595640.html#p10292236. 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: toughandtrickyexponentsandrootsquestions12595640.html#p10292247. 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: toughandtrickyexponentsandrootsquestions12595640.html#p10292278. 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: toughandtrickyexponentsandrootsquestions12595640.html#p10292289. 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: toughandtrickyexponentsandrootsquestions12595640.html#p102922910. Given that \(5x=1253y+z\) and \(\sqrt{5x}5\sqrt{z3y}=0\), then what is the value of \(\sqrt{\frac{45(z3y)}{x}}\)?A. 5 B. 10 C. 15 D. 20 E. Can not be determined Solution: toughandtrickyexponentsandrootsquestions12595640.html#p102923111. 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: toughandtrickyexponentsandrootsquestions12595640.html#p1029232
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NEW!!! Tough and tricky exponents and roots questions [#permalink]
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28 Sep 2017, 02:30
Bunuel wrote: 5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a nonnegative 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 \(n=0\) > \(n^{26}26^n=0^{26}26^0=01=1\). 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. hi man 26^n is a multiple of an odd number, this is only possible when n = 0, but why?, yes it can be equal to zero, but "n" can also be a nonnegative irrational number... Is this because the base to which the power "n" is raised is even or anything else ...? please say to me. thanks in advance, man



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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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28 Sep 2017, 02:34
gmatcracker2017 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 nonnegative 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 \(n=0\) > \(n^{26}26^n=0^{26}26^0=01=1\). 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. hi man 26^n is a multiple of an odd number, this is only possible when n = 0, but why?, yes it can be equal to zero, but "n" can also be a nonnegative irrational number... thanks in advance, man Check the highlighted part.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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28 Sep 2017, 02:42
Bunuel wrote: gmatcracker2017 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 nonnegative 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 \(n=0\) > \(n^{26}26^n=0^{26}26^0=01=1\). 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. hi man 26^n is a multiple of an odd number, this is only possible when n = 0, but why?, yes it can be equal to zero, but "n" can also be a nonnegative irrational number... thanks in advance, man Check the highlighted part. oh! man I got it thanks a lot man



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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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01 Oct 2017, 16:16
Bunuel wrote: SOLUTIONS:
1. What is the value of \(\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}}\)? 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 unsquare 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 \((xy)^2=x^22xy+y^2\)).
So we get: \((\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})+(\sqrt{2510\sqrt{6}})^2=\) \(=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})+(2510\sqrt{6})\).
Note that sum of the first and the third terms simplifies to \((25+10\sqrt{6})+(2510\sqrt{6})=50\), so we have \(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})\) > \(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(2510\sqrt{6})}\).
Also must know for the GMAT: \((x+y)(xy)=x^2y^2\), thus \(50+2\sqrt{(25+10\sqrt{6})(2510\sqrt{6})}=50+2\sqrt{25^2(10\sqrt{6})^2)}=50+2\sqrt{625600}=50+2\sqrt{25}=60\).
Recall that we should unsquare this value to get the right the answer: \(\sqrt{60}=2\sqrt{15}\).
Answer: C. Hi BunuelI found it easier to do the ''math'' 6^1/2 = 2,45 (just know by heart) So first term would be 25+10*2,45 = a little bit more than 49, so the square root is 7,something small Second term would be 2510*2,45 = almost 1, just considered 1, and than the square root = 1 little bit more than 7  1 > answer should be something close to 6 A: 2 * square root(5) = 2*2,43 , not close to 6 B: Sroot(55) = bigger than 7 C: Looks good D: No way E: No way Choice is C.



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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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08 Nov 2017, 13:46
Bunuel wrote: 2. What is the units digit of \((17^3)^41973^{3^2}\)? 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 1113=2, which gives the final answer that the units digit of \((17^3)^41973^{3^2}\) is 2.
Answer B. I don't understand the last step why is it not 113? which =7?



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08 Nov 2017, 20:20
arcticTO wrote: Bunuel wrote: 2. What is the units digit of \((17^3)^41973^{3^2}\)? 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 1113=2, which gives the final answer that the units digit of \((17^3)^41973^{3^2}\) is 2.
Answer B. I don't understand the last step why is it not 113? which =7? Please reread the highlighted part.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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14 Feb 2018, 08:40
Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}\)? 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.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. 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



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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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14 Feb 2018, 08:47
gmatmo wrote: Bunuel wrote: 3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}\)? 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.
\(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\)
\(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\)
Answer: E. 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/rulesforpo ... l#p1096628
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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