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 GRE.

$a^1=a$ 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 $(a^m)^n$

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:
$a^{\frac{1}{n}}=\sqrt[n]{a}$

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

Exponential Equations:
When solving equations with even exponents, we must consider both positive and negative possibilities for the solutions.

For instance $a^2=25$, the two possible solutions are $5$ and $-5$.

When solving equations with odd exponents, we'll have only one solution.

For instance for $a^3=8$, solution is $a=2$ and for $a^3=-8$, solution is $a=-2$.

Exponents and divisibility:
$a^n-b^n$ is ALWAYS divisible by $a-b$.
$a^n-b^n$ is divisible by $a+b$ if $n$ is even.

$a^n + b^n$ is divisible by $a+b$ if $n$ is odd, and not divisible by a+b if n is even.


LAST DIGIT OF A PRODUCT

Last $n$ digits of a product of integers are last $n$ digits of the product of last $n$ digits of these integers.

For instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60

Example: The last digit of 85945*89*58307=5*9*7=45*7=35=5?


LAST DIGIT OF A POWER

Determining the last digit of $(xyz)^n$:

1. Last digit of $(xyz)^n$ is the same as that of $z^n$;
2. Determine the cyclicity number $c$ of $z$;
3. Find the remainder $r$ when $n$ divided by the cyclisity;
4. When $r>0$, then last digit of $(xyz)^n$ is the same as that of $z^r$ and when $r=0$, then last digit of $(xyz)^n$ is the same as that of $z^c$, where $c$ is the cyclisity number.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. $(xy4)^n$) have a cyclisity of 2. When n is odd $(xy4)^n$ will end with 4 and when n is even $(xy4)^n$ will end with 6.
• Integers ending with 9 (eg. $(xy9)^n$) have a cyclisity of 2. When n is odd $(xy9)^n$ will end with 9 and when n is even $(xy9)^n$ will end with 1.

Example: What is the last digit of $127^{39}$?
Solution: Last digit of $127^{39}$ is the same as that of $7^{39}$. Now we should determine the cyclisity of $7$:

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!)
...

So, the cyclisity of 7 is 4.

Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of $127^{39}$ is the same as that of the last digit of $7^{39}$, is the same as that of the last digit of $7^3$, which is $3$.


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 GRE 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 GRE.

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

• For GRE it's good to memorize following values:
$\sqrt{2}\approx{1.41}$
$\sqrt{3}\approx{1.73}$
$\sqrt{5}\approx{2.24}$
$\sqrt{6}\approx{2.45}$
$\sqrt{7}\approx{2.65}$
$\sqrt{8}\approx{2.83}$
$\sqrt{10}\approx{3.16}$