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EXPONENTS & ROOTS

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, walker

--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
[read more] [AppStore]

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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 \((-3)^0=1\)

\(a^1=a\) 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.


Perfect Square

A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square.

There are some tips about the perfect square:
• The number of distinct factors of a perfect square is ALWAYS ODD.
• The sum of distinct factors of a perfect square is ALWAYS ODD.
• A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
• Perfect square always has even number of powers of prime factors.


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 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 \(\sqrt[3]{-64} =-4\).

• For GMAT it's good to memorize following values:
\(\sqrt{2}\approx{1.41}\)
\(\sqrt{3}\approx{1.73}\)
\(\sqrt{5}\approx{2.24}\)
\(\sqrt{6}\approx{2.45}\)
\(\sqrt{7}\approx{2.65}\)
\(\sqrt{8}\approx{2.83}\)
\(\sqrt{10}\approx{3.16}\)
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This is a lucky day for me :-). Thank you. +1Kudos.
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Great resource. Thanks people!
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good stuff. thank you!
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FRACTIONS

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, walker

--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
[read more] [AppStore]

--------------------------------------------------------

Definition

Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.

Fraction can be expressed in two forms fractional representation \((\frac{m}{n})\) and decimal representation \((a.bcd)\).


Fractional representation

Fractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole).

• The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, \(\frac{9}{7}\), 9 is the numerator and 7 is denominator.

• Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. \(\frac{1}{3}\) is a proper fraction.

• Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. \(\frac{5}{2}\) is improper fraction.

• An integer combined with a proper fraction is called mixed number. \(4\frac{3}{5}\) is a mixed number. This can also be written as an improper fraction: \(\frac{23}{5}\)


Converting Improper Fractions

• Converting Improper Fractions to Mixed Fractions:
1. Divide the numerator by the denominator
2. Write down the whole number answer
3. Then write down any remainder above the denominator
Example #1: Convert \(\frac{11}{4}\) to a mixed fraction.
Solution: Divide \(\frac{11}{4} = 2\) with a remainder of \(3\). Write down the \(2\) and then write down the remainder \(3\) above the denominator \(4\), like this: \(2\frac{3}{4}\)

• Converting Mixed Fractions to Improper Fractions:
1. Multiply the whole number part by the fraction's denominator
2. Add that to the numerator
3. Then write the result on top of the denominator
Example #2: Convert \(3\frac{2}{5}\) to an improper fraction.
Solution: Multiply the whole number by the denominator: \(3*5=15\). Add the numerator to that: \(15 + 2 = 17\). Then write that down above the denominator, like this: \(\frac{17}{5}\)


Reciprocal

Reciprocal for a number \(x\), denoted by \(\frac{1}{x}\) or \(x^{-1}\), is a number which when multiplied by \(x\) yields \(1\). The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\). To get the reciprocal of a number, divide 1 by the number. For example reciprocal of \(3\) is \(\frac{1}{3}\), reciprocal of \(\frac{5}{6}\) is \(\frac{6}{5}\).


Operation on Fractions

Adding/Subtracting fractions:

To add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.

To add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions

Multiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.

Dividing fractions: Change the divisor into its reciprocal and then multiply.

Example #1: \(\frac{3}{7}+\frac{2}{3}=\frac{9}{21}+\frac{14}{21}=\frac{23}{21}\)

Example #2: Given \(\frac{\frac{3}{5}}{2}\), take the reciprocal of \(2\). The reciprocal is \(\frac{1}{2}\). Now multiply: \(\frac{3}{5}*\frac{1}{2}=\frac{3}{10}\).


Decimal Representation

The decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....).

Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).


Converting Decimals to Fractions

• To convert a terminating decimal to fraction:
1. Calculate the total numbers after decimal point
2. Remove the decimal point from the number
3. Put 1 under the denominator and annex it with "0" as many as the total in step 1
4. Reduce the fraction to its lowest terms

Example: Convert \(0.56\) to a fraction.
1: Total number after decimal point is 2.
2 and 3: \(\frac{56}{100}\).
4: Reducing it to lowest terms: \(\frac{56}{100}=\frac{14}{25}\)

• To convert a recurring decimal to fraction:
1. Separate the recurring number from the decimal fraction
2. Annex denominator with "9" as many times as the length of the recurring number
3. Reduce the fraction to its lowest terms

Example #1: Convert \(0.393939...\) to a fraction.
1: The recurring number is \(39\).
2: \(\frac{39}{99}\), the number \(39\) is of length \(2\) so we have added two nines.
3: Reducing it to lowest terms: \(\frac{39}{99}=\frac{13}{33}\).

• To convert a mixed-recurring decimal to fraction:
1. Write down the number consisting with non-repeating digits and repeating digits.
2. Subtract non-repeating number from above.
3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's.

Example #2: Convert \(0.2512(12)\) to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=2487;
3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.


Rounding

Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.

Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.


Ratios and Proportions

Given that \(\frac{a}{b}=\frac{c}{d}\), where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.

\(\frac{b}{a}=\frac{d}{c}\) - invertendo

\(\frac{a}{c}=\frac{b}{d}\) - alternendo

\(\frac{a+b}{b}=\frac{c+d}{d}\) - componendo

\(\frac{a-b}{b}=\frac{c-d}{d}\) - dividendo

\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\) - componendo & dividendo

Thanks, this a great and concise summary of all the basics.
Just to add, another important basic rule


for +ve fractions,
if \(\frac{a}{b}<1\) and x is any positive number then \(\frac{(a+x)}{(b+x)}>\frac{a}{b}\)
,in short when the two fractions have same difference between numerator and denominator , the fraction having greater numerator is always greater.

Also, when \(\frac{a}{b}>1\) and x is any positive number then \(\frac{(a+x)}{(b+x)}<\frac{a}{b}\)


Example which one is greater of the following two:
a)\(\frac{111487}{111490}\) b)\(\frac{111587}{111590}\) , the ans is b for the same reasons above, as \(\frac{111487+100}{111490+100}\)
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• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form 6n−1 or 6n+1, because all other numbers are divisible by 2 or 3.

Why does this property "all prime numbers above 3 are of the form 6n−1 or 6n+1" hold true for 961, but 961 is not a prime? This holds true for square of any prime number? Is there any gap in my understanding?
6n+1 = 961 --> 6n = 960
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Very useful...keep posting..
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Very useful...keep posting..

Thank you.

For more check ALL YOU NEED FOR QUANT ! ! !.

Hope it helps.
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Thanks so much! Super useful!! :-D

Question: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?

Many thanks!
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Bunuel
Thanks so much! Super useful!! :-D

Question: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?

Many thanks!

2. Properties of Integers



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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