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655-705 Level|   Algebra|   Exponents|   Roots|      
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IanStewart
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NEW!!! Tough and tricky exponents and roots questions 12 Jan 2012, 14:31
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LalaB
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And it looks as though every other question has been solved except for this one. I won't solve it completely, but I'll help people get started. Here, since we are asked for an approximation only, we don't need to compute an exact value. You might notice that if x is a positive integer, \(22^{22x}\) is vastly bigger than \(22^{2x}\). Since that's true, if we only need to estimate, we can ignore the small term. We can do the same thing in the denominator.

2a^(2ax)/a^(ax)=2*a^(ax)=2*11^(11x)

ans is
Show Spoiler
B
?
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Your substitution might be making things more difficult. If we want to estimate the value of

\(\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}\)

then we can ignore the comparatively small values in the numerator and denominator; this is roughly equal to


\(\frac{22^{22x}}{11^{11x}}\)

which is equal to

\(\frac{22^{22x}}{11^{11x}} = \frac{2^{22x} 11^{22x}}{11^{11x}} = 2^{22x} 11^{11x}\)
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NEW!!! Tough and tricky exponents and roots questions 13 Jan 2012, 21:05
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Q1.

\(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)

First look and we know that this is of the type:

let a = \(\sqrt{25+10\sqrt{6}}\)

Let b = \(\sqrt{25-10\sqrt{6}}\)

let \(a+b=x\)

Then \((a+b)^2=x^2\)

So \(a^2+b^2+2ab=x^2\)

So \(a^2 = {25+10\sqrt{6}}\)
So \(b^2 = {25-10\sqrt{6}}\)
So \(a^2+b^2=50\)

Now on to \(2ab\)

\(a=\sqrt{25+10\sqrt{6}}\)
\(b=\sqrt{25-10\sqrt{6}}\)

let \(ab=y\)

then \(\sqrt{25+10\sqrt{6}}*\sqrt{25-10\sqrt{6}}=y\)
So \(({25+10\sqrt{6}})*({25-10\sqrt{6}})=y^2\)
so \(625-600=y^2\)
so \(y=5\)
so \(2ab=2y=10\)
so \(x^2=a^2+b^2+2ab=50+10=60\)
so \(x^2=60\)
so \(x=\sqrt{4*15}\)
so \(x=2\sqrt{15}\)

Hence Answer is C. \(2\sqrt{15}\)
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NEW!!! Tough and tricky exponents and roots questions 13 Jan 2012, 21:45
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Q3.

\(5^{10x}=4900\)
so \({(5^{5x})}^2=70^2\)

Take square root of both sides >

so \(5^{5x}=70\)

Also from question stem:

\(2^{\sqrt{y}} = 5^2\)

So (from the question stem)

\(4^{-\sqrt{y}} = \frac{1}{4^{\sqrt{y}}}\)
\(4^{-\sqrt{y}} = (\frac{1}{2^{\sqrt{y}}})^2 = \frac{1}{5^4}\)

Now \(5^{(x-1)^5}\)

Should be \(\frac{5^{5x}}{5^5}\)

We know \(5^{10x}=4900\)

so \(5^{10x}=70^2\)

so \(5^{5x}=70\)

So \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)

is basically \(\frac{5^x}{5^5}*5^4\)

We already know the value of \(5^x\) which is \(70\)

So now it becomes \(\frac{70}{5^5}*5^4\)

Which should resolve to \(14\)

Hence Answer = E = \(14\)
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NEW!!! Tough and tricky exponents and roots questions 13 Jan 2012, 21:46
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Q4. What is the value of \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5\)

This one was actually the simplest I thought. Here is how:

\(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5\)

So \(5*(1+4)+4*5^2+4*5^3+4*5^4+4*5^5\)
So \(5^2+4*5^2+4*5^3+4*5^4+4*5^5\)
So \(5^2*(1+4)+4*5^3+4*5^4+4*5^5\)
So \(5^3+4*5^3+4*5^4+4*5^5\)
So \(5^3*(1+4)+4*5^4+4*5^5\)

So every expression behind contributes a power of 1 to the one in front of it. We just need to see the last which is \(5^5\)
Keep solving and you come to a total of \(5^6\)

Hence Answer = \(5^6\)
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NEW!!! Tough and tricky exponents and roots questions 15 Jan 2012, 21:04
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“The only true wisdom is in knowing you know nothing.”
― Socrates


back to studying!
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NEW!!! Tough and tricky exponents and roots questions 06 Mar 2012, 23:53
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Quote:
What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8
Show more



Quote:

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 note that the second number is mush 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.
Show more


I agree with all the steps above but i have this doubt in this

unit digit of 1st number is 1.
Unit digit of second number is 3

1-3
would it not mean
unit digit is 8?
for eg if we have
551 -853 ..Unit digit will be 2
but what if it is 551-453 .unit digit will be 8?

so how do we know which is the larger number of these, although it says X-Y?
Please let me know,.
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NEW!!! Tough and tricky exponents and roots questions 07 Mar 2012, 01:36
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Quote:
What is the units digit of (17^3)^4-1973^{3^2}?
A. 0
B. 2
C. 4
D. 6
E. 8



Quote:

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.


I agree with all the steps above but i have this doubt in this

unit digit of 1st number is 1.
Unit digit of second number is 3

1-3
would it not mean
unit digit is 8?
for eg if we have
551 -853 ..Unit digit will be 2
but what if it is 551-453 .unit digit will be 8?

so how do we know which is the larger number of these, although it says X-Y?
Please let me know,.
Show more

That's a little trap there.

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.

So, as you can see the cases like 551-453 (larger number minus smaller number) are not possible.

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

Hope it's clear.
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Bunuel
4^{\sqrt{y}
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For Question #3.
I'm confused why 4 squareroot Y = (2 squareroot Y)^2 instead of (2 ^2 squareroot Y).

Not sure I understand the properties correctly...
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NEW!!! Tough and tricky exponents and roots questions 23 Apr 2012, 13:21
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KGG88
Bunuel
4^{\sqrt{y}

For Question #3.
I'm confused why 4 squareroot Y = (2 squareroot Y)^2 instead of (2 ^2 squareroot Y).

Not sure I understand the properties correctly...
Show more


\(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2\).

Check this for properties: tough-and-tricky-exponents-and-roots-questions-125956.html#p1027888

Hope it helps.
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NEW!!! Tough and tricky exponents and roots questions 25 Jun 2012, 09:59
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Bunuel
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.


Hey Bunuel,

I understood the logic behind finding the unit places, but how can u determine if the reminder is 2 or 8

in this example
(17^12) - (1973^9) = unit place is 2 (agreed- i actually did the calculation using calculator ;) )
because : (17^12) unit place = 1
(1973^9) unit place = 3

but if we have just - (7^12) - (3^9) or (3^9) - (7^12) we have unit place as "8"

3^9 = 19683 - unit place will still be 2
7^12 = 13841287201 - unit place will still be 1

so my question is how will u determine if the answer is 2/8? clearly we have 2 different answers??
am i missing anything??
Show more

The units digit of \(17^{12}-1973^{9}\) is 2 and not 8 since \(1973^{9}\) is much larger number than \(17^{12}\), thus their difference will be negative, something like 11-13=-2.

If we had something like 21-13 (if the first number were greater than the second one), then the units digit of their difference would be 8.

Hope i's clear.
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NEW!!! Tough and tricky exponents and roots questions Updated on: 15 Oct 2012, 01:35
Originally posted by EvaJager on 29 Jun 2012, 11:34.
Last edited by EvaJager on 15 Oct 2012, 01:35, edited 1 time in total.
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Bunuel
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.
Show more

Another way to do it, using the same formulas:

\(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{5(5+2\sqrt{6})}+\sqrt{5(5-2\sqrt{6})}=\sqrt{5}\sqrt{5+2\sqrt{6}}+\sqrt{5}\sqrt{5-2\sqrt{6}}=\)
\(=\sqrt{5}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=\sqrt{5}(\sqrt{(\sqrt{3}+\sqrt{2})^2}+\sqrt{(\sqrt{3}-\sqrt{2})^2})=\sqrt{5}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=\sqrt{5}(2\sqrt{3})=2\sqrt{15}\)
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Bunuel
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

Given: \(x>0\) and \(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\) --> \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), as the expression under the square root extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\). Square both sides: \(x^2=6+x\) --> \((x+2)(x-3)=0\) --> \(x=-2\) or \(x=3\), but since \(x>0\) then: \(x=3\).

Answer: B.
Show more

Hi Bunuel - All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate?

\(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), as the expression under the square root extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\).

Cheers
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Bunuel
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

Given: \(x>0\) and \(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\) --> \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), as the expression under the square root extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\). Square both sides: \(x^2=6+x\) --> \((x+2)(x-3)=0\) --> \(x=-2\) or \(x=3\), but since \(x>0\) then: \(x=3\).

Answer: B.

Hi Bunuel - All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate?

\(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), as the expression under the square root extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\).

Cheers
Show more

Given: \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\). Consier the expression in brackets: \((\sqrt{6+\sqrt{6+\sqrt{6+...}}})\). It's the same as the right hand side of the initial expression, thus it also equals to \(x\). When replaced we'll have: \(x=\sqrt{6+x}\).

Hope it's clear.
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Thanks Bunuel for these interesting problems.

With respect to problem number 2

"So, we have that the units digit of is 1 and the units digit of 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 is 2."

I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number

eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.

I am unable to think any other way. Please let me know where I am going wrong.

Thanks
gmatrant
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gmatrant
Thanks Bunuel for these interesting problems.

With respect to problem number 2

"So, we have that the units digit of is 1 and the units digit of 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 is 2."

I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number

eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.

I am unable to think any other way. Please let me know where I am going wrong.

Thanks
gmatrant
Show more

First of all 757,571 - 58,299,374,483 = -58,298,616,912

You could consider easier cases: 11-13=-2, 11-23=-12, 21-83=-62, ...
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You talk about even and odd in this problem. Wouldn't this not be enough? Since 26 has factors of 2 and 13, even if you had added some even numbers in the product it would still not have sufficed? You would need a 13, 39 etc ? Is my reasoning correct?

Bunuel
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.
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NEW!!! Tough and tricky exponents and roots questions 31 Mar 2013, 08:33
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jeremystaub28
You talk about even and odd in this problem. Wouldn't this not be enough? Since 26 has factors of 2 and 13, even if you had added some even numbers in the product it would still not have sufficed? You would need a 13, 39 etc ? Is my reasoning correct?

Bunuel
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.
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The fact that the product is odd, is already enough to say that it cannot be a multiple of 26^n unless n=0.
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ranaazad
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NEW!!! Tough and tricky exponents and roots questions 10 Jul 2015, 17:55
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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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ranaazad
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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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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Hope it helps.
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Bunuel
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.
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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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