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NEW!!! Tough and tricky exponents and roots questions [#permalink]
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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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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. 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. Is this approach correct? Many thanks,



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08 Dec 2014, 03:10
joaogallegomoura 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. 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. Is this approach correct? Many thanks, ______________ Yes, that's correct.
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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? Thanks.
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11 Jul 2015, 02:40



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29 Jun 2016, 11:07
OMG .. there is a much elegant and easier method to solve this \(\sqrt{6} = ~2.4 ==> 10\sqrt{6}=10*2.4==> 24\) 25+24=49 \(\sqrt{49} =7\) 2524=1 \(\sqrt{1} = 1\) So our question is 7+1 = 8 See option \(2 \sqrt{15} = 2*4= 8\) HENCE C 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. OMG .. there is a much elegant and easier method to solve this \(\sqrt{6} = ~2.4 ==> 10\sqrt{6}=10*2.4==> 24\) 25+24=49 \(\sqrt{49} =7\) 2524=1 \(\sqrt{1} = 1\) So our question is 7+1 = 8 See option \(2 \sqrt{15} = 2*4= 8\) HENCE C
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14 Feb 2018, 09:40
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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14 Feb 2018, 09:47
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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08 Apr 2018, 10:57
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. ======================================================================================================================== Hi Bunuel I have a doubt here. If we are doing this subtraction (13). And, if we would have actually solved the whole subtraction of the number. the we would have borrowed 10 from the tens digit. So, the final subtraction would have looked like (113)= 8 please let me know if I am correct or not. if not, then why?



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08 Apr 2018, 11:35
shivangibh 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. ======================================================================================================================== Hi Bunuel I have a doubt here. If we are doing this subtraction (13). And, if we would have actually solved the whole subtraction of the number. the we would have borrowed 10 from the tens digit. So, the final subtraction would have looked like (113)= 8 please let me know if I am correct or not. if not, then why? You could very easily test cases to prove your logic. (positive number ending with 1)  (greater number ending with 3) = (negative number ending with 2) 1123=12 31133=102 41123=82 .... I tried to explain this several times in this thread: https://gmatclub.com/forum/newtoughan ... l#p1054715https://gmatclub.com/forum/newtoughan ... l#p1099223In addition here is an actual result: \((17^3)^41973^{3^2}=453047530293560259230589943252\)
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12 Apr 2018, 23:51
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 ALTERNATIVE SOLUTION.Let \(\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}}\)=x. Then \(x^2\)=(\(\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}})^2\) And \(x^2\)=\(25+10\sqrt{6}+2\sqrt{625600}+25\sqrt{6}\) \(x^2=60\) => x=\(2\sqrt{15}\) Hence C.
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