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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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New post 07 Dec 2014, 10:20
1
1
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.

Is this approach correct?

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

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New post 08 Dec 2014, 02:10
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 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.

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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New post 10 Jul 2015, 16:55
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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New post 11 Jul 2015, 01:40
1
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:
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which-of-the-following-is-closest-to-10180-1030-a-110224.html
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if-x-10-10-x-2-2x-7-3x-2-10x-2-is-closest-to-143897.html
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which-of-the-following-is-closest-in-value-to-64425.html
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if-x-3-000-then-the-value-of-x-2x-1-is-closest-to-166128.html

Hope it helps.
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NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 29 Jun 2016, 10:07
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

\(\sqrt{49} =7\)

25-24=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 \(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)?
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.



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
\(\sqrt{49} =7\)
25-24=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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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 14 Feb 2018, 08:40
Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}\)?
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.

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

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

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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New post 14 Feb 2018, 08:47
1
gmatmo wrote:
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.

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

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

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/rules-for-po ... l#p1096628
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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

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New post 08 Apr 2018, 09:57
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.


========================================================================================================================

Hi Bunuel

I have a doubt here.
If we are doing this subtraction (1-3). 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 (11-3)= 8
please let me know if I am correct or not. if not, then why?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 08 Apr 2018, 10:35
1
shivangibh 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.


========================================================================================================================

Hi Bunuel

I have a doubt here.
If we are doing this subtraction (1-3). 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 (11-3)= 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)

11-23=-12
31-133=-102
41-123=-82
....

I tried to explain this several times in this thread:
https://gmatclub.com/forum/new-tough-an ... l#p1054715
https://gmatclub.com/forum/new-tough-an ... l#p1099223

In addition here is an actual result:

\((17^3)^4-1973^{3^2}=-453047530293560259230589943252\)
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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

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New post 12 Apr 2018, 22:51
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

ALTERNATIVE SOLUTION.
Let \(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)=x.
Then \(x^2\)=(\(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2\)
And \(x^2\)=\(25+10\sqrt{6}+2\sqrt{625-600}+25-\sqrt{6}\)
\(x^2=60\) => x=\(2\sqrt{15}\)
Hence C.
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