THEORY TO TACKLE THE PROBLEMS ABOVE:
For more on number theory check the Number Theory Chapter of Math Book: math-number-theory-88376.html

EXPONENTS

Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number a multiplied n times can be written as a^n, where a represents the base, the number that is multiplied by itself n times and n represents the exponent. The exponent indicates how many times to multiple the base, a, by itself.

Exponents one and zero:
a^0=1 Any nonzero number to the power of 0 is 1.
For example: 5^0=1 and (-3)^0=1
• Note: the case of 0^0 is not tested on the GMAT.

a^1=a Any number to the power 1 is itself.

Powers of zero:
If the exponent is positive, the power of zero is zero: 0^n = 0, where n > 0.

If the exponent is negative, the power of zero (0^n, where n < 0) is undefined, because division by zero is implied.

Powers of one:
1^n=1 The integer powers of one are one.

Negative powers:
a^{-n}=\frac{1}{a^n}

Powers of minus one:
If n is an even integer, then (-1)^n=1.

If n is an odd integer, then (-1)^n =-1.

Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
a^n*b^n=(ab)^n

\frac{a^n}{b^n}=(\frac{a}{b})^n

(a^m)^n=a^{mn}

a^m^n=a^{(m^n)} and not (a^m)^n (if exponentiation is indicated by stacked symbols, the rule is to work from the top down)

Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
a^n*a^m=a^{n+m}

\frac{a^n}{a^m}=a^{n-m}

Fraction as power:
a^{\frac{1}{n}}=\sqrt[n]{a}

a^{\frac{m}{n}}=\sqrt[n]{a^m}


ROOTS

Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.

General rules:
• \sqrt{x}\sqrt{y}=\sqrt{xy} and \frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}.

• (\sqrt{x})^n=\sqrt{x^n}

• x^{\frac{1}{n}}=\sqrt[n]{x}

• x^{\frac{n}{m}}=\sqrt[m]{x^n}

• {\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}

• \sqrt{x^2}=|x|, when x\leq{0}, then \sqrt{x^2}=-x and when x\geq{0}, then \sqrt{x^2}=x

• When the GMAT provides the square root sign for an even root, such as \sqrt{x} or \sqrt[4]{x}, then the only accepted answer is the positive root.

That is, \sqrt{25}=5, NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

• Odd roots will have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4.
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
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. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


What are GMAT Club Tests?
25 extra-hard Quant Tests

Find out what's new at GMAT Club - latest features and updates

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

7^12-3^9
11-3=8

ANS IS E
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I get 2. for 1st term 1 and for the 2nd term 9. 1st term - 2nd term= unit digit=2

Edited
What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
= 5+4*5[1+5+5^2+5^3+5^4]
=5+20*781
=15625
=625*25=5^6

Can you explain 1? some of the code came out messed up and its difficult to read
_________________

Reward wisdom with kudos

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

1st raise it to the 2nd power (in a simple form u have (a+b)^2)
(\sqrt{25+ 10[square_root]6}[/square_root]+\sqrt{25- 10[square_root]6}[/square_root] )^2 =

=((\sqrt{25+ 10[square_root]6}[/square_root])^2+ 2((\sqrt{25+ 10[square_root]6}[/square_root])*(\sqrt{25- 10[square_root]6}[/square_root] ) +(\sqrt{25- 10[square_root]6}[/square_root] )^2 (in a simple form u have got (a^2+2ab+b^2)

=25+ 10\sqrt{6}[/square_root]+25-10\sqrt{6}[/square_root] +2((\sqrt{25+ 10[square_root]6}[/square_root])*(\sqrt{25- 10[square_root]6}[/square_root]))

=50+2*\sqrt{625-6*100}=50+2*5=60

\sqrt{60}=2\sqrt{15}

hope I could be helpful
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Q 5:

2^n*13^n * K= 23^2*5^8*3^18*29^8 [ after prime factorization]
value of n has to be 0 for 2^n*13^n to be 1
therefore, n^2*6 -26^n
=0-1
=-1
(??)

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*29^8/26^n= 23^2*25^4*27^6*29^8/(2*13)^n from this point it is obvious, that n =0 (since both 2 and 13 are primes and none of the numbers of numerator can be divisible by 2 and 13)

n=0 then n^26-26^n=0^26-26^0=0-1=-1

answ is C
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

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)

x^2=2^{64}
x^2=(2^{32})^2 so x =2^32 from now I dont know ...either C, or E is the answ. I would choose C

@Bunuel, when r u going to post answers? just curious
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

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


Units digit for 17^3 = 3
Units digit for (17^3)^4 = 3^4 = 1
Units digit for 1973^9 = 3
Units digit for xxx1 - xxxx3 = 8

Option