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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]
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.

HOW the second number is much larger then the first one?
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
BelalHossain046 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.

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/new-tough-an ... l#p2043010

Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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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 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.

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

Yes, x is NOT a multiple of 13, bur 26^n IS if n > 0, so n must be 0.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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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

5+4*5+4*5^2+4*5^3+4*5^4+4*5^5
= 5 + (5 - 1)*5 + (5 - 1)*5^2 + (5 - 1)*5^3 + (5 - 1)*5^4 + (5 - 1)*5^5
= 5 + (5^2 - 5) + (5^3 - 5^2) + (5^4 - 5^3) + (5^5 - 5^4) + (5^6 - 5^5)

Observe that alternate terms cancel out, giving us:

Sum = 5^6

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
Bunuel wrote:
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 $$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}$$?
A. $$2\sqrt{5}$$
B. $$\sqrt{55}$$
C. $$2\sqrt{15}$$
D. 50
E. 60

Solution: https://gmatclub.com/forum/tough-and-tri ... l#p1029216

2. What is the units digit of $$(17^3)^4-1973^{3^2}$$?
A. 0
B. 2
C. 4
D. 6
E. 8

Solution: https://gmatclub.com/forum/tough-and-tri ... l#p1029219

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

Solution: https://gmatclub.com/forum/tough-and-tri ... l#p1029221

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

Solution: https://gmatclub.com/forum/tough-and-tri ... l#p1029222

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

Solution: https://gmatclub.com/forum/tough-and-tri ... l#p1029223

6. 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: https://gmatclub.com/forum/tough-and-tri ... l#p1029224

7. 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: https://gmatclub.com/forum/tough-and-tri ... l#p1029227

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

Solution: https://gmatclub.com/forum/tough-and-tri ... l#p1029228

9. 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: https://gmatclub.com/forum/tough-and-tri ... l#p1029229

10. Given that $$5x=125-3y+z$$ and $$\sqrt{5x}-5-\sqrt{z-3y}=0$$, then what is the value of $$\sqrt{\frac{45(z-3y)}{x}}$$?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

Solution: https://gmatclub.com/forum/tough-and-tri ... l#p1029231

11. 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: https://gmatclub.com/forum/tough-and-tri ... l#p1029232

A question about chained exponents. In the second question you say:

$${1973^{3}}^{2} = 1973^{9}$$

But, as the '3' digit is not bigger in font size than the '2' digit I interpreted it as:

$${1973^{3}}^{2} = 1973^{6}$$

and my interpretacion is consistent with the way we write mathematical formulas here:

{1973^{3}}^{2} (look at the braces)

For example, if you have:

$${{{{7^{2}}^{3}}^{4}}^{5}}$$

How should we interpret it?

There are a lot of different interpretations and results.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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andreagonzalez2k wrote:
A question about chained exponents. In the second question you say:

$${1973^{3}}^{2} = 1973^{9}$$

But, as the '3' digit is not bigger in font size than the '2' digit I interpreted it as:

$${1973^{3}}^{2} = 1973^{6}$$

and my interpretacion is consistent with the way we write mathematical formulas here:

{1973^{3}}^{2} (look at the braces)

For example, if you have:

$${{{{7^{2}}^{3}}^{4}}^{5}}$$

How should we interpret it?

The font size actually doesn't matter. The only thing that matters is whether there are parentheses. When there are no parentheses, as in your examples, we always apply the exponents from the top down. So

$$\\ 7^{2^5} = 7^{32}\\$$

(with no brackets, we always interpret it to mean $$7^{(2^5)}$$) and is not equal to

$$\\ (7^2)^5 = 7^{10}\\$$

You actually don't see the first situation about very often on the GMAT -- it's rare (but not impossible) that you'll see something like

$$\\ 2^{3^2}\\$$

(which equals 2^9) in a GMAT question. Usually when I've seen something this situation on the GMAT, there's an unknown in the exponent, e.g.:

$$\\ 5^{x^2}\\$$

(which is not the same thing as $$5^{2x}$$.) The situation with parentheses, though, is one that comes up all the time when you're solving exponent questions, so it's likely what you've encountered most often.

And the number you ask about, $${{{{7^{2}}^{3}}^{4}}^{5}}$$, might look harmless enough, but it is absurdly enormous. Just the 4^5 part is already roughly 1000, then we raise 3 to that power to get something close to 10^500. Then we need to raise the 2 to that power, and raise the 7 to whatever absurdly large number we get. That's not something you'd ever need to deal with on the GMAT, fortunately.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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: $$\sqrt{60}=2\sqrt{15}$$.

Shouldn't the answer be +2sqrt15 or -2sqrt15? An equation with a squared variable always has two solutions. Bunuel TargetTestPrep
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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HoudaSR wrote:
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: $$\sqrt{60}=2\sqrt{15}$$.

Shouldn't the answer be +2sqrt15 or -2sqrt15? An equation with a squared variable always has two solutions. Bunuel TargetTestPrep

So, we get that $$(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2=\sqrt{60}$$.

$$\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}$$ is the sum of two square roots, which cannot be negative.

Mathematically, $$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root.

The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.

Hope it helps.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
please add more questions like this... not only in this session but please add this type of questions in other topics like fractions, profits etc...
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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pudu wrote:
please add more questions like this... not only in this session but please add this type of questions in other topics like fractions, profits etc...

Check the list below:

Hope it helps.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
Q7. This is actually very easy. Notice that, 3^2 is 9, 3^3 is 2.
First term > 3^2 i.e. 9, so we can say that the root is obviously greater than 3. Same goes with the next term, 3√9 > 2^3, so it is greater than 2 aswell. For the rest of the terms, they are all greater than 1. If we add every element, it will always be greater than 12.
Hence, option E

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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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 [b]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.[/b]

In the last line I did Units digit for xxx1 - xxxx3 = 8, Can you help what am i doing wrong.
[Reference -You mentioned unit digit will be 2 as 1-2= -2]
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
GMATBUDDING 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 [b]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.[/b]

In the last line I did Units digit for xxx1 - xxxx3 = 8, Can you help what am i doing wrong.
[Reference -You mentioned unit digit will be 2 as 1-2= -2]

I think your doubt is addressed in the solution you quite. Here it is again in more details:

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 $$c^n$$.

Hence, the units digit of $$(17^3)^4$$ is the same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is the same as that of $$3^{3^2}$$.

II. $${(a^m)}^n=a^{mn}$$ and $$a^{m^n}=a^{(m^n)}$$.

Hence, $${(7^3)}^4=7^{(3*4)}=7^{12}$$ and $$3^{3^2}=3^{(3^2)}=3^9$$.

III. The units digit of integers in positive integer powers repeats in a specific pattern called cyclicity. The units digit of 7 and 3 in positive integer powers repeats in patterns of 4:

1. $$7^1=7$$(last digit is 7)

2. $$7^2=49$$(last digit is 9)

3. $$7^3=xx3$$(last digit is 3)

4. $$7^4=xxx1$$ (last digit is 1)

5. $$7^5=xxxxxx7$$ (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!)

...

Hence, the units digit of $$17^{12}$$ is 1, as the 12th number in the repeating pattern 7-9-3-1 is 1, and the units digit of $$1973^9$$ is 3, as the 9th number in the repeating pattern 3-9-7-1 is 3.

BACK TO THE QUESTION:

Thus, we know that the units digit of $${(17^3)}^4=17^{12}$$ is 1, and the units digit of $$1973^{3^2}=1973^9$$ is 3. If the first number is greater than the second number, the units digit of their difference will be 8, for example, 11 - 3 = 8. However, if the second number is greater than the first number, the units digit of their difference will be 2, for example, 11 - 13 = -2.

Consider this, even if the first number were $$100^{12}$$ instead of $$17^{12}$$, and the second number were $$1000^9$$ instead of $$1973^9$$, the first number, $$100^{12}=10^{24}$$, would still be smaller than the second number $$1,000^9=10^{27}$$. Therefore, $$17^{12} < 1973^9$$, and the units digit of the difference is 2.

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

Hi, can anyone explain why we need to un-square the value? What is the rule or logic behind this?

Thank you.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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dendenden wrote:
Quote:
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:

Hi, can anyone explain why we need to un-square the value? What is the rule or logic behind this?

Thank you.

Hello dendenden

Below is detailed answer. I hope it helps!
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
1
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dendenden wrote:
Quote:
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:

Hi, can anyone explain why we need to un-square the value? What is the rule or logic behind this?

Thank you.

Squaring an expression containing square roots is a common algebraic technique used to eliminate the square roots and simplify the expression. Conversely, taking the square root is an operation performed to balance the initial squaring. Essentially, first we square to simplify and eliminate the square roots but at the end to we get the squared value of the expression, and to get the original value, we need to unsquare.

Below is more detailed solution, which might help:

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

First, square the given expression to eliminate the square roots, but remember to take the square root of the result at the end to balance the operation and obtain the correct answer.

Important for the GMAT: $$(x+y)^2=x^2+2xy+y^2$$ and $$(x-y)^2=x^2-2xy+y^2$$.

Following these rules, 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 the sum of the first and third terms simplifies to $$(25+10\sqrt{6})+(25-10\sqrt{6})=50$$, so we have:

$$50+2(\sqrt{25 + 10\sqrt{6} })(\sqrt{25 - 10\sqrt{6} })=$$

$$=50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6}) }$$.

Another important concept for the GMAT: $$(x+y)(x-y)=x^2-y^2$$. Using this, we can simplify further:

$$50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})}=$$

$$=50+2\sqrt{25^2-(10\sqrt{6})^2)} =$$

$$= 50+2\sqrt{625-600}=$$

$$=50+2\sqrt{25}=$$

$$=60$$.

Finally, remember to take the square root of this value to obtain the correct answer: $$\sqrt{60}=2\sqrt{15}$$.