NUMB3RS
NUMB3RS
Concept 1: Categorization of numbers Concept 2: Recurring decimals as rational numbers Real
Imaginary
Let
x = 0.77777… (1)
then 10x = 7.77777… (2) Rational
Irrational
Subtracting (1) from (2)… 9x = 7 So x = = 0.7777….
Non-recurring/non-terminating
Natural numbers
Prime
Whole numbers
Integers
Similarly 0.7272727272 =
Fractions
Composite
Concept 3: Prime & Composite numbers Proper
Improper
A prime number is a number that is divisible by 1 and itself only. Eg.2, 7, 17, 101 etc.
Mixed
Decimal
Vulgar
(Please note 1 is neither a prime number nor a composite number) (The only even prime number is 2) A composite number is one, which is a product of one or more primes. Eg.6, 9, 42 etc.
Definitions: Imaginary number: square root of any negative number Eg. √- 4 = √i24 = ± i√2 (where i2 = -1)
Concept 4: Factorials Real Number: Any number that can be marked on the number line. Rational Number: Any number that can be written in the form of p/q where q ≠ 0.
Represented as n! = n.(n-1).(n-2).(n-3)…3.2.1 Eg. 6! = 6.5.4.3.2.1 = 720 and 0! = 1 Factorials play an important role in permutation & combination.
Irrational number: defined as non-recurring and non-terminating eg. Root of any prime number Natural number N: N є 1, 2, 3……∞ Concept 5: The number line Whole numbers W: W є 0, 1, 2, 3….. ∞ Integers I :I є -∞…-3, -2, -1, 0, 1, 2, 3….∞
A number line is a line with 0 at the center and numbered from 1 to ∞ at equal intervals towards the right of 0 and from -1 to -∞ towards the left of 0. -∞
Proper fraction: , Improper fraction: , Mixed fraction: 1 , -3
(denominator > numerator) (numerator > denominator) (combination of integer and a proper fraction)
-3
-2
-1
0
1
2
3
∞
NUMB3RS
NUMB3RS
Concept 6: Absolute value Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies.
Concept 7: Brackets in interval notation x (5,∞) means x lies between 5 and ∞ not including 5 or ∞
The absolute value of a number is never negative.
x (-∞, 4] U [8, 14] means lies between -∞ to 5 not including -∞ but including 5 and between 8 to 14 including both
It is called absolute or mod value. Mod x or absolute value of x is denoted as IxI Eg. IxI = 5. Means x = ±5 IxI > 5 means -5 > x > 5 IxI < 5 means -5 < x < 5 Q:
Concept 8: Prime factors of a number The prime factors of a positive integer are the prime numbers that divide that integer exactly, without leaving a remainder.
If I2x-7I = 5. Find x
DIVISIBILITY TESTS Divisibility Tests
Eg. 84 can be divided by the prime numbers 2, 3 and 7. So the prime factors of 84 are 2,3 and 7. Q: What are the prime factors of 3465. Example
A number is divisible by 2 if the last digit is 0, 2, 4, 168 is divisible by 2 since the last digit is 6 or 8. 8. A number is divisible by 3 if the sum of the digits is 168 is divisible by 3 since the sum of the divisible by 3. digits is 1+6+8=15, and 15 is divisible by 3. A number is divisible by 4 if the last 2 digits of the number are divisible by 4.
316 is divisible by 4 since 16 is divisible by 4.
A number is divisible by 5 if the last digit is either 0 195 is divisible by 5 since the last digit is or 5. 5. A number is divisible by 6 if it is divisible by 2 AND 168 is divisible by 6 since it is divisible it is divisible by 3. by 2 AND it is divisible by 3. A number is divisible by 7 if the number obtained 182 is divisible by 7 since 18 + 5x2 = 28 by adding5 times the number of units to the tens is is divisible by 7. divisible by 7 A number is divisible by 8 if the last 3 digits of the number are divisible by 8.
7,120 is divisible by 8 since 120 is divisible by 8.
A number is divisible by 9 if the sum of the digits is 549 is divisible by 9 since the sum of the divisible by 9. digits is 5+4+9=18, and 18 is divisible by 9. A number is divisible by 10 if the last digit is 0.
x [5,∞) means x lies between 5 and ∞ including 5 but not including ∞
1,470 is divisible by 10 since last digit is 0.
A number is divisible by 11 if the difference 1738 is divisible by 11 since (7+8) – between the sum of the digits in the odd place and (1+3) = 11 is divisible by 11. the sum of the digits in the even place = 0 or a multiple of 11
Concept 9: Factors of a number If N = am x bn, where a & b are prime factors Then the number of factors of N = (m+1).(n+1) Eg. 84 = 22 x 3 x 7 No. of factors of 84 = (2+1)(1+1)(1+1) = 12 (Perfect squares have odd number of factors. The reverse is also true. If an integer has odd number of factors then the integer is a perfect square Q: Find the number of odd factors of 6300 HCF & LCM Concept 10: HCF of numbers (HCF of a set of integers must be ≤ the smallest number of the set) The highest common factor (HCF) of two or more numbers is the largest number that is a factor of all of the given numbers. HCF is also known as GCD – Greatest Common Denominator Eg. The HCF of 40, 48 and 56 40 = 2 x 2 x 2 x 5 48 = 2 x 2 x 2 x 2 x 3 56 = 2 x 2 x 2 x 7 The common factors for all 3 numbers are 2 x 2 x 2 = 8 Alternately: Find the HCF of (48-40) and (56-48)
Q:Find the HCF of 48, 84 and 132
NUMB3RS Q:The maximum number of children amongst whom 1001 pens and 910 pencils can be distributed such that each student gets the same number of pens and the same number of pencils?
NUMB3RS The LCM of fractions = Eg.
Concept 11: LCM of numbers (LCM of a set of integers must be ≥ the largest number of the set)
Find LCM of
and
.
LCM of 126 and 90 = 630 and HCF of 15 and 25 = 5 So LCM of
and
is
= 126
The least common multiple (LCM of two or more integers, is the smallest positive integer that is divisible by all the relevant integers Eg. The LCM of 40, 48 and 56 48 = 2 x 2 x 2 x 2 x 3 56 = 2 x 2 x 2 x 7 80 = 2 x 2 x 2 x 2 x 5 LCM = Product of (Prime factors common to three xPrime factors common to two x Remaining prime factors LCM = (2 × 2 × 2) × (2) × (3 × 5 × 7)
COMPARISONS BETWEEN NUMBERS a> b then, a + c > b + c and a –c > b – c a> b and c > 0 then, ac>bc and a/c > b/c if a, b ≥ 0 and n > 0 an>bn and 1/an< 1/bn
Concept 12: Relation between HCF &LCM of two numbersa & b. Product of a and b = HCF(a,b) x LCM(a,b)
Ix – yI = Iy – xI Ix.yI = IxI. IyI
Concept 13: HCF of Fractions Ix + yI ≤ IxI + IyI
The HCF of fractions =
Ix + yI ≥ IxI - IyI Eg.
Find HCF of
and
.
HCF of 126 and 102 = 6 and LCM of 15 and 25 = 75
INDICES & SURDS Concept 15: Rules of Indices
So HCF of
and
=
am × bm = (a.b)m
am × an = a(m+n)
am ÷ an = a(m-n)
(am)n = amn
am ≠(am)n
Eg.
= 28whereas (22)3 = 26
Concept 16: Surds Concept 14: LCM of Fractions
A surd is an irrational number. All square roots of prime numbers are surds.
NUMB3RS
√2 is a surd. √2 = 21/2
Cube root of 4 is written as √ = 4
√a .√b = √(ab)
1/3
NUMB3RS Units digit of n
Units digit of n, n2, n3, n4, n5, ….
Cycle
1
1,1,1…
1
2
2, 4, 8, 6, 2, 4, ….
4
Complex surd – combination of a rational and irrational term Eg. 7 + 3√2
3
3, 9, 7, 1, 3, 9, ……
4
Conjugate of a surd
4
4, 6, 4, 6, ……..
2
5
5, 5, 5, ……
1
6
6, 6, 6, ……..
1
7
7, 9, 3, 1, 7, 9, ……
4
Eg. √2 ×√13 = √26
Eg. If 7 + 3√2 is a surd, its conjugate is 7 - 3√2
Concept 17: Rationalization of Surds If a surd is of the form
√ √
the concept of rationalization is to make the
8
8, 4, 2, 6, 8, 4, …..
4
denominator a rational number. So the numerator and the denominator are
9
9, 1, 9, 1, …..
2
multiplied by the conjugate of the denominator.
0
0, 0, 0……
1
√
Eg.
√
=
√
√
√
√
√ –
{(a+b)(a-b) = a2 – b2}
√ √
=
√
The unit digit is 7 and the cycle for 7 is 4 i.e. 254 is 63 complete cycles and the 2nd part of the 64th cycle
√
√
The unit digit of the 2nd part of a cycle for n = 7 is 9 So unit digit of 37254 is 9
√
=
Eg. Find the units digit of 37254
So 254 = 4×63 + 2
–
=
=
=
√
√
Q: Rationalize
√
√
√
√
Concept 21: Rule of cyclicity for Remainders Eg. Find the remainder when 4326 ÷ 7
Concept 18: Square root of quadratic surds 2
A quadratic surd is of the form (√a + √b) = a + b + 2√ab
41/7 R = 4, 44/7 R = 4,
42/7 R = 2,
43/7 R = 1
45/7 R = 2,
46/7 R = 1
Eg. (√2 + √3)2 = 2 + 3 + 2√2√3 = 5 + 2√6
Cycle for 4n/7 is 3
Conversely if a surd is of the form a + b + 2√ab, then its sq. root is √a + √b.
326 = 3×108 + 2
Eg. 7 + 4√3 = 7 + 2√(4.3) = √(7 + 2√12) = √4 + √3 = 2 + √3
i.e 326 = 108 complete cycles and the 2nd part of the 109th cycle
Similarly √(7 - 4√3) = 2 -√3
So the remainder when 4326 ÷ 7 is 2
Q: Find the square root of 8 - √60
Concept 19: Rule of Cyclicity for unit’s digit
The unit’s digit of successive power’s of n follow a cycle
PROPERTIES OF NUMBERS
NUMB3RS Concept 22: The product of n consecutive integers is divisible by n!
NUMB3RS an – bn is divisible by a + b for n even an + bn is divisible by a + b for n odd
p
Concept 23: n – n is always divisible by p (where n is a whole number and p is a prime) Q: What’s the remainder when 1511 is divided by 11?
an + bn is not divisible a - b for any n
REMAINDER THEOREM & REMAINDERS Concept 24:
Concept 25:
Odd + Odd = Even Odd + Even = Odd Even + Even = Even
Odd x Odd = Odd Odd x Even = Even Even x Even = Even
If a function of x f(x) is divided by x-a then the remainder R is a function of ‘a’ i.e. R = f(a) Eg.x3 – 3x2 + 7x -8 is divided by x-2 Then R = f(a) = 23 – 3(2)2 + 7.2 – 8 = 2 Also if R = 0 then x-2 is a factor of f(x) (If the sum of the coefficients of the function = 0 then x-1 is a factor)
ALGEBRAIC RELATIONSHIPS
(If the sum of the odd powers of x = sum of even powers of x, then x+1 is a factor)
Concept 26: Some common formulae
Eg. If 222 x 224 x 226 is divided by 13 what is the remainder.
(a + b)(a - b) = a2 – b2
When 222 is divided by 13 the remainder is 1
(a + b)2 = a2 + 2ab + b2
When 224 is divided by 13 the remainder is 3
(a - b)2 = a2 – 2ab + b2
When 226 is divided by 13 the remainder is 5
(a + b)3 = a3 +3ab(a + b) + b3
So the problem can be treated as 1 x 3 x 5 ÷ 13 = 15 ÷ 13
(a - b)3 = a3 – 3ab(a - b) – b3
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Q:
What is the value of 182 + 172 – 162 - 152
Remainder = 2 Eg. If 225 + 227 + 229 is divided by 13 what is the remainder. When 225 is divided by 13 the remainder is 4
Concept 27: a3 + b3 + c3 -3abc = (a + b + c)(a2 + b2 + c2- ab – bc - ca) So a3 + b3 + c3 = 3abc if a + b + c = 0
When 227 is divided by 13 the remainder is 6 When 229 is divided by 13 the remainder is 8 So the problem can be treated as (4 +6 +8) ÷ 13 = 18 ÷ 13 Remainder = 5
Concept 28: Divisibility an – bn is divisible by a - b for all n
NUMB3RS BINOMIAL THEOREM n
n
n
n
n-1
Concept 30: Other number bases 1
n
.
(a + b) = C0.a + C1 .a .b + C2 a
n-2
2
n
1
.b +……. Cn-1.a .b
n-1
n
0
+ Cn.a .b
n
Also an – bn = (a – b)(an-1.b0 +an-2.b1 + an-3.b2+ …….a1.bn-2 + a0.bn-1) Eg. What is the 40th term of (2x+y)42 40th term = 42C39.(2x)3.y39 40th term = 11,480.23.x3..y39 40th term = 91840 x3.y39 Eg. What is the remainder when 1717 is ÷ 6 1717 = (18-1)17 Expanding 1817 + 17C1.1816.(-1)1+ 17C21815.(-1)2 +….17C16.181.(-1)16 + (-1)17 Now, all the terms except the last term is divisible by 6, so the remainder is (-1)17 = -1 hence positive remainder = 6 – 1 = 5
NUMBER BASED SYSTEMS Decimal system:
NUMB3RS
The decimal system has 10 as a base The decimal system has 10 digits i.e. digits 0 to 9
Concept 29: Mathematical notation of a number A number xyz in a decimal system denotes x.102 + y.101 + z.100 xyz = 100x + 10y + z
The base number is always written as a subscript either as 5648 or (564)8.
A base of n has digits 0 to n-1 i.e an octal system has digits 0 to 7.
Base is 12, digits are 0, 1, 2…9, A, B (A denotes 10 and B denotes 11)
Concept 31: Conversion from decimal to other bases A number in the decimal system can also be obtained by arranging the remainders obtained by successive division of the number by 10 as below: 10
372
R
10
37
2
10
3
7
0
3
(372)10 is converted to another system (say, octal – 8) in the same manner 8
372
R
8
46
4
8 8
5 0
6 5
= (564) 8 = (372)10 (Note: As the base decreases the number value increases)
Concept 32: Addition and subtraction in other bases When 2 or more digits are added, what is in excess of base/multiples of base is written and the base/multiples of base carried to the next higher place. Eg. (4 7 6)8 + (5 7 7)8 (1 2 7 5)8
Eg. 372 = 3.102 + 7.101 + 2.100 = 300 + 70 + 2 All number based systems follow the same notation. (256)8 denotes the number 256 to the base 8 and (256)8 = 2.82 +5.81 + 6.80 = 128 + 40 + 6 = (174)10
= 372
Unit’s,
6 + 7 = 13, excess of base is 13 - 8 =5 write 5 carry 1
Ten’s,
1+7+7=15 excess of base 15–8=7 write 7 carry 1
Hundred’s,
1+4+5=10 excess of base 10-8=2 write 2 carry 1
Thousand’s, 1 write 1. Concept 33: Multiplication When two numbers are multiplied what is in excess of base/multiples of base is written and the base/multiples of base carried to the next higher place.
NUMB3RS Eg.
(4 7) 8 × (5 7) 8 421 173 ( 2 3 5 1) 8
Concept 34: Decimal to Binary conversion Binary system (Base 2) – Digits 0 and 1 Conversion from decimal to binary – Methodology Eg.
Find 237 to base 2, 4 and 8
Find the nearest power of 2 larger than the number 28 = 256
28 – 1 in binary = 11111111 ( eight 1’s) = 255
To obtain 237, subtract 255-237 = 18 from the above i.e 16+2
1 1 1 1 1 1 1 1 = 111 0 1 1 0 1 = 23710
From base 2 to base 4, take digits 2 at a time (11)(10)(11)(01) = 32214
From base 2 to 8, take digits 3 at a time (11)(101)(101) = 355 8
NUMB3RS
