Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 27 Oct 2014
Posts: 8

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
07 Dec 2014, 11:20
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,



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
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.
_________________



Manager
Joined: 16 Jan 2013
Posts: 74
Location: Bangladesh
GMAT 1: 490 Q41 V18 GMAT 2: 610 Q45 V28
GPA: 2.75

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
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.
_________________
Heading towards perfection>>



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
11 Jul 2015, 02:40
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. Similar questions to practice: thevalueof108102107103isclosesttowhichof95082.htmlwhichofthefollowingisclosestto101801030a110224.htmlwhichofthefollowingbestapproximatesthevalueofqif99674.htmlnewtoughandtrickyexponentsandrootsquestions12595640.htmlifx1010x22x73x210x2isclosestto143897.html10000100410thevalueoftheexpressionaboveis59398.htmlwhichofthefollowingisclosestinvalueto64425.html101801030whichofthefollowingbestapproximates84309.htmlifx2b8886forwhichofthefollowingbvalues160197.htmlifx3000thenthevalueofx2x1isclosestto166128.htmlHope it helps.
_________________



Director
Joined: 04 Jun 2016
Posts: 547

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
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
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE : 17th SEPTEMBER 2016. .. 16 March 2017  I am back but for all purposes please consider me semiretired.



Intern
Joined: 02 Dec 2017
Posts: 16

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



Math Expert
Joined: 02 Sep 2009
Posts: 59587

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



Intern
Joined: 27 May 2017
Posts: 5
GPA: 4

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



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
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: \((17^3)^41973^{3^2}=453047530293560259230589943252\)
_________________



Manager
Joined: 28 Nov 2017
Posts: 137
Location: Uzbekistan

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
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.
_________________



Intern
Joined: 10 Feb 2018
Posts: 10
GMAT 1: 670 Q44 V38 GMAT 2: 710 Q48 V40
WE: Project Management (Health Care)

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
15 Dec 2018, 10:49
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. Can you please post a solution using Usubstitution. Replacing 10sqrt{6} with U from the beginning. I think I would lose track of all these numbers on a test. Thanks!



Manager
Joined: 27 Oct 2017
Posts: 72

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
07 Jan 2019, 20:29
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. I think the answer should be E ..Can you please check... as 50+2*sqrt25 >50+10>60..



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
07 Jan 2019, 21:14
arorni wrote: 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. I think the answer should be E ..Can you please check... as 50+2*sqrt25 >50+10>60.. Check the highlighted parts in the solution. We need to take the square root from 60 to get the final answer.
_________________



Intern
Joined: 17 Jun 2019
Posts: 9
Location: United States
GPA: 3.92

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
20 Aug 2019, 03:05
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: \(sum=\frac{b*(r^{n}1)}{r1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\).
So in our case: \(5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^51)}{51})=5^6\).
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. My approach is like that: 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5= 5(4+1)+4*5^2+4*5^3+4*5^4+4*5^5= 5^2 +4*5^2+4*5^3+4*5^4+4*5^5= 5^2(4+1)+4*5^3+4*5^4+4*5^5= 5^3(4+1)+4*5^4+4*5^5= 5^4(1+4)+5^5=5^5(1+4)=5^6 Answer A



Manager
Joined: 11 Feb 2013
Posts: 229
Location: United States (TX)
GMAT 1: 490 Q44 V15 GMAT 2: 690 Q47 V38
GPA: 3.05
WE: Analyst (Commercial Banking)

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
02 Sep 2019, 00:40
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. HOW the second number is much larger then the first one?



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
02 Sep 2019, 00:51
BelalHossain046 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. HOW the second number is much larger then the first one? 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. I explained this several times in this thread: https://gmatclub.com/forum/newtoughan ... l#p2043010Hope it's clear.
_________________



Intern
Joined: 19 Jan 2019
Posts: 3
Location: India
Concentration: Technology, Leadership
GPA: 4

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
03 Oct 2019, 02:53
My approach was 26^n is possible with only multiples of 2 and 13. 13 is a prime number and not in the given value of x. thus,n has to zero. Is this correct ? 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.



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
03 Oct 2019, 02:58
r3shabh wrote: My approach was 26^n is possible with only multiples of 2 and 13. 13 is a prime number and not in the given value of x. thus,n has to zero. Is this correct ? 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. Yes, x is NOT a multiple of 13, bur 26^n IS if n > 0, so n must be 0.
_________________



Manager
Status: In last prep stage
Joined: 11 Jun 2017
Posts: 167
GMAT 1: 630 Q44 V33 GMAT 2: 680 Q47 V37
GPA: 3.2

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
29 Oct 2019, 00:37
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. Great methodology used by Bunuel! But, I could not go on this thought process and my mind got all blank seeing this question and thinking of a way to solve this. I could not think of any algebraic method to solve this within 2 min. I have learned and applied all the great exponent rules mentioned above by Bunuel but in this case, could not start the steps, so I could not use the exponent formulas Can the forum please provide some other method or solution to this. chetan2u, @puspitk



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
Show Tags
29 Oct 2019, 00:43
StudiosTom 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. Great methodology used by Bunuel! But, I could not go on this thought process and my mind got all blank seeing this question and thinking of a way to solve this. I could not think of any algebraic method to solve this within 2 min. I have learned and applied all the great exponent rules mentioned above by Bunuel but in this case, could not start the steps, so I could not use the exponent formulas Can the forum please provide some other method or solution to this. chetan2u, @puspitk You can check the post two above yours for alternative method.
_________________




Re: NEW!!! Tough and tricky exponents and roots questions
[#permalink]
29 Oct 2019, 00:43



Go to page
Previous
1 2 3 4 5 6
Next
[ 103 posts ]



