Aptitude and Maths

Quotes by Shaurya(me)

Complete the thing that you wanted to complete in the first place because you may not get a second chance to do it again/from start.

Number System

Any Composite number can be written as the product of its prime factors.
Eg: 240 = 2⁴ 3¹ 5¹
Eg: N = a^p q^q r^r
Factors of 240 are (2⁰,2¹,2²,2³,2⁴)(3⁰,3¹)(5⁰,5¹).
Sum of factors of 240 are (2⁰ + 2¹ + 2² + 2³ + 2⁴)(3⁰ + 3¹)(5⁰ + 5¹).
Total #factors(TF) = #primeFactors + #compositeFactors + 1(NPNC)
#factors of 120 = 3 + 12 + 1 = 16
Total #factors(TF) = (p+1)*(q+1)*(r+1) = 522 = 20
Total Number of Even Factors(e) = (p)*(q+1)*(r+1)
Total Number of odd Factors = TF - e
Product of Factors
n^(TF/2) = 240^(20/2) = 240¹⁰.
n^(TF/2) = 36^(9/2) = 36^4.5.

Unit place

Unit place of (1! + 2! + 3! +...... + 99!) = Unit place of (1 + 2 + 6 + 24 + 120 + 720 + ...) = 3.
Last two digits of (1! + 2! + 3! +...... + 99!) = Last two digits of (1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 + 3628800 + ...) = 73.

Cyclicity

Cyclicity of 4:-

2 : -> 2(4n+1) 4(4n+2) 8(4n+3) 6(4n) -> 2 4 8 6 -> 2 4 8 6 -> 2 4 8 6 -> 2 4 8 6 ->....
3 : -> 3(4n+1) 9(4n+2) 7(4n+3) 1(4n) -> 3 9 7 1 -> 3 9 7 1 -> 3 9 7 1 -> 3 9 7 1 ->....
7 : -> 7(4n+1) 9(4n+2) 3(4n+3) 1(4n) -> 7 9 3 1 -> 7 9 3 1 -> 7 9 3 1 -> 7 9 3 1 ->....
8 : -> 8(4n+1) 4(4n+2) 2(4n+3) 6(4n) -> 8 4 2 6 -> 8 4 2 6 -> 8 4 2 6 -> 8 4 2 6 ->....

Cyclicity of 2:-

4 : -> 4(2n+1) 6(2n) -> 4 6 -> 4 6 -> 4 6 -> 4 6 ->....
9 : -> 9(2n+1) 1(2n) -> 9 1 -> 9 1 -> 9 1 -> 9 1 ->....

Cyclicity of 1:-

0 : -> 0 -> 0 -> 0 ...
1 : -> 1 -> 1 -> 1 ...
5 : -> 5 -> 5 -> 5 ...
6 : -> 6 -> 6 -> 6 ...

HCF/GCD and LCM

(a,b) are Co-Prime if gcd(a,b) = 1.
- Used in the calculation of MMI in Number Theory.
(a,b) are Twin-Prime if both a and b are prime and b-a = 2.
- used in the calculation of MMI in Number Theory.
HCF and LCM of fractions
- HCF(a/b, c/d, e/f,...) = HCF(a,c,d)/LCM(b,d,f) = HCM(numerator)/ LCM(denominator)
- LCM(a/b, c/d, e/f,...) = LCM(a,c,d)/HCF(b,d,f) = LCM(numerator)/ HCF(denominator)
HCF of two numbers will always be less than the LCM of them
- HCF(a,b) < LCM(a,b) ----- Always.
Product of two numbers is equal to the product of LCM and HCF of them
- ab = HCF(a,b) LCM(a,b)
- Ques 1: What is the product of two numbers whose HCF is 25 and LCM is 5.
  - Ans: Not Possible, beacuse HCF > LCM
- Ques 2: What is the product of two numbers whose HCF is 10 and LCM is 35.
  - Ans: 10*35 = 350
Remainder Theorem
- Let P(x) be a polynomial. If P(x) is divided by (x-a), then the Remainder is P(a).
  - Ex: find the Remainder when (x² + x - 1) is divisible by (x-5).
    - Ans: P(5) = 25 + 5 - 1 = 29.
Factor Theorem
- If P(x) is divisible by (x-a), then P(a) = 0.
  - In other words, If P(x) is divisible by (x - a), then Remainder = P(a) = 0.
- Note: when we divide P(x) by f(x), the Remainder will always be of atleast 1 degree less than the f(x) Polynomial.
Remainder after dividing Questions:-
- Find the number that leaves remander 0 after dividing by 12 , 15 both.
  - Case: Remainder 0
  - Ans: k*LCM(12,15) = 60k; where k = 1,2,3,4,....
- Find the number that leaves remander 7(r) after dividing by 12 , 15 both.
  - Case: Remainder same
  - Ans: k*LCM(12,15) + r = 60k + 7; where k=1,2,3,4,......
- Find the number that leaves remander 4, 7 after dividing by 12 , 15 respectively.
  - Case: diff = (Divisor - remainder) is same
  - Ans: k*LCM(12,15) - diff = 60k - 8
- Find the number that leaves remander 7, 13 after dividing by 12 , 15 respectively.
  - Case: Other/rest
  - Let number be N
  - N = 12p + 7 = 15q + 13
  - BY hit and trial we get p = 3, q = 2
  - N = k*LCM(12,15) + Least Value of N = 60k + 43

🌟 IIT Level Tips 🌟

underroot/square root("√") is a +ve quantity => √16 = 4. and not ±4.
√x^2 = |x| = mod x = ±x
√f(x) exist only when f(x) ≥ 0
x^2, x^3, x^4, ... are called power functions.
- Their values are always +ve and can be zero. i.e. x^a ≥ 0.
2^x, 3^x, 4^x, ... are called exponential functions, and their values are always +ve and can never be zero.

=> a^x > 0.

If sum of two or more +ve quantities results in zero, that implies all quantities are zero.

Ex. Given: √x + |u| + v^2 = 0.
=> x = 0, u = 0, v = 0

Quadratic Equation

ax^2 + bx + c > 0
- If a > 0 (Upward parabola) & D < 0 (no real roots)
ax^2 + bx + c < 0
- If a < 0 (Downward parabola) & D < 0 (no real roots)

Formula: (a^4 + a^2 + 1) = (a^2 + a + 1)(a^2 - a + 1)
Solving Inequalites

1. Factorize the given equation and find all odd powers bracket
1. Get values from bracket and arrange them in number line.
1. Find values of x for which function is not defined (even power bracket or denominator) and remove/take them.
1. Put +ve sign to the rightmost side and then change of sign for odd powers brackets only.
1. chose answer
- a. For > 0, >= 0, opt +ve sign
- b. For < 0, <= 0, opt -ve sign

a^3 + b^3 +c^3 - 3abc

= (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac)
= (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)/2

Every perfect square number is of the form:

4n or 4n+1
3n or 3n+1
2n(either even) or 2n+1(or odd) 😂

If LCM(a,b) = c, then c/a and c/b are both Integers

LCM(n!, (n+1)!) = (n+1)!
LCM(6,9) = 18
LCM(pi, pi^2) = Not possible, because if you think answer is pi^2, then pi^2/pi and pi^2/pi^2 should have been both Integers, but they are not.
LCM(√3, √6) = Not possible, same reason

Every Prime Number above 3 are of the form

6n + 1 or 6n - 1(≅ 6n + 5)

Ratio and Proportion

Ratio is a:b or a/b
Compound Ratio
- If a/b and c/d are two ratios then (ac)/(bd) is a Compound ratio.
Proportion
- if a/b and c/d are two ratios and they are like a/b = c/d then, both ratios are said to be in proportion.
- written as a:b :: c:d, where b and c are means & a and d are extremes.
- Product of means = Product of extremes.
Continued-Proportion
- If a,b,c are in continued-proop. then a:b :: b:c => b^2 = ac
- => any three consecutive terms of a GP are in continued Proportion.
🌟 IIT Level Notes 🌟
- C & D
  - If(a/b) = c/d --- (1)
  - then a/b + 1 = c/d + 1 is called Compendendo
  - then a/b - 1 = c/d - 1 is called Dividendo
  - and doing both is called C and D.
- If a/b = c/d = e/f = ... ---(2)
- expresion (2) = any linear Combination of Numerators / same linear Combination of Denominators
- Example a/b = c/d = e/f = (2a + 3c + 5e)/ (2b + 3d + 5f)
- Example a/b = c/d = e/f = [(a^(1/n) + c^(1/n) + e^(1/n))/ (b^(1/n) + d^(1/n) + f^(1/n))]^n
- proof: express (2) = k and put values to validate (k-Method)

Logarithms

If loga b = x, then
required 3 Conditions should hold: a>0, a!=1, b>0
If b=0, loga b = loga 0, = -∞ (if a>1, let a=2, then 2^(-∞) = 0) = Not defined.
If b=0, loga b = loga 0, = ∞ (if a<1, let a=0.5, then (0.5)^(∞) = 0) = Not defined.
If a=1,b=1, loga b = log1 1, = R(many answers) = Not defined.
- logcot45 tan45 = Not defined as base cannot be 1.
Properties of log(may not be known to you)
- Smile/muskan Theroem: loga b^c = logc b^a, if you draw the curve line b/w swapped elements, you will see a smile.
- a^x = e^(xlna)
- loga b^c = c*loga b
- loga^c b = (1/c)(loga b) = base ki power 1 upon ho jayegi.
Logarithms of negative numbers is not defined. See the log graph.
Properties
- log 1(base a) = 0 for all a > 0 and a != 1
- log a(base a) = 1 for all a > 0 and a != 1
Number of digits in a number can be found by taking the log base 10 of that number.
- floor(log99(base 10) + 1) = 2.
- floor(log100(base 10) + 1) = 3.
VVVI Imp:
- In loga b,
- If both a and b are on same side of unity, then answer is +ve, else -ve
- Ex - log(2) (4) = +ve = 2
- Ex - log(2) (.4) = -ve = log2 (4/10) = log2 4 - log2 10 = 2 - 3.something = -ve
- Ex - log(.2) (4) = -ve = log(2/10) 4 = log2 4/log2 (2/10) = log2 4/(log2 2 - log2 10) = Denom is negative, num is +ve = -ve
- Ex - log(.2) (.4) = +ve
- In other words
  - If loga b > 0
    - => a>1 and b>1
    - or => 0<a<1 and 0<b<1
  - If loga b < 0
    - => a > 1 and 0 < b < 1
    - or => 0 < a < 1 and b > 1

Progression

🌟 IIT Level Notes 🌟
- AP
  - last term = a + (#terms-1)*d
  - => number of terms(n) in AP = (l-a)/d + 1
  - Tn means nth term
  - Tn = a + (n - 1)d = a+nd-d
    - => Tn is a linear function
    - => coefficient of n is d.
  - Arithmatic Mean = (a+b)/2
  - 😈😈 Inserting Arithmetic Means between numbers
    - Let a & b are two numbers and we want to insert n means b/w them
    - Total Numbers in the end will be n + 2.
    - Let means be Arithmatic means. since these n+2 numbers must form an AP with first term as a and last being b.
    - last term = a + (#terms-1)*d
    - b = a + (n + 1)*d
    - d = (b-a)/(n+1)
    - => first term = a
    - => second term (1st AM) = a+d = a + (b-a)/(n+1)
    - => third term (2st AM) = a+2d = a + 2*(b-a)/(n+1)
    - and so on...
- GP
  - Geometric Mean = sqrt(ab)
  - 😈😈 Inserting Geometric Means between numbers
    - Let a & b are two numbers and we want to insert n means b/w them
    - Total Numbers in the end will be n + 2.
    - Let means be Geometric means. since these n+2 numbers must form an GP with first term as a and last being b.
    - last term = ar^(#terms-1)
    - b = ar^(n + 1)
    - r = (b/a)^(1/(n+1))
    - => first term = a
    - => second term (1st GM) = ar = a(b/a)^(1/(n+1))
    - => third term (2st GM) = ar^2 = a((b/a)^(1/(n+1)))^2
    - and so on...
- HP
  - There is no general formulas for HP. THey are found by inverting the numbers and then using AP formulas.
- 😈😈 Relation b/w AM,GM,HM
  - GM is the Geometric mean of AM and HM
    - 😈 GM = sqrt(AM*HM)
  - For any two numbers a &b
    - 😈 AM > GM > HM (Trick: Lexiographically decreasing)
- AGP

Time and Distance

Relative Speed
- If two cars are moving toward/away from each other(in opposite directions), the relative speed is S1+S2
- If two cars are moving in the same direction, the relative speed is |S1-S2|
Trains
- case 1: Train crossing stationary object, and has no length
  - s(train) = l(train length) / time
- case 2: Train crossing stationary object, and has length
  - s(train) = l+o(train length + object length) / time
- case 3: Train crossing moving object, and has no length
  - case 3a: Same direction.
    - |s(train) - s(object)| = l(train length) / time
  - case 3a: opposite direction.
    - s(train) + s(object) = l(train length) / time
- case 4: Train crossing moving object, and has length
  - case 3a: Same direction.
    - |s(train) - s(object)| = l+o(train length + object length) / time
  - case 3a: opposite direction.
    - s(train) + s(object) = l+o(train length + object length) / time
- Conclusion: lengths(train and object) will always be added.
  - All cases combined are covered in case 4. just put the length or speed of object 0 to get other cases.
Boats and Stream
- upstream :- Boat Moving opposite to the direction of stream. => speed of the boat will decrease.
  - Boat relative Speed : s(boat) - s(stream)
- downstream:- Boat Moving with stream => speed of the boat will increase.
  - Boat relative Speed : s(boat) + s(stream)

Time and Work

Time is always told in hr, sec, min {hours, minutes, seconds}.
Work is always referred to as efficiency => How much thing is done in this much time.

Data Interpretation

bar graph
line graph
histogram
polygon graph
tabular form
pie chart

Deductive Logic Problems

Deriving a conclusion from single/multiple propositions is called premises.

Functions

Limits

Limit is defined at a point, if both LHL and RHL are defined and Equal.
7 Indeterminant Forms:-
- ∞/∞, 0/0, 0*∞, ∞ - ∞, 1^∞, 0⁰, ∞⁰
Solving these 7 Interminant forms:-
- ∞/∞, 0/0, 0*∞, => Use Hospital's Rule
- ∞ - ∞ => use factorization to convert to ∞/∞, 0/0, 0*∞ forms.
- 1^∞ => for lim f(x)^g(x), use formula e^{lim g(x) * [f(x) - 1]}
- 0⁰ or ∞⁰ => take log both side and solve.

Aptitude and Maths

Quotes by Shaurya(me)​

Number System​

Unit place​

Cyclicity​

Cyclicity of 4:-​

Cyclicity of 2:-​

Cyclicity of 1:-​

HCF/GCD and LCM​

🌟 IIT Level Tips 🌟​

Ratio and Proportion​

Logarithms​

Progression​

Time and Distance​

Time and Work​

Data Interpretation​

Deductive Logic Problems​

Functions​

Limits​

Continuity​

Quotes by Shaurya(me)

Number System

Unit place

Cyclicity

Cyclicity of 4:-

Cyclicity of 2:-

Cyclicity of 1:-

HCF/GCD and LCM

🌟 IIT Level Tips 🌟

Ratio and Proportion

Logarithms

Progression

Time and Distance

Time and Work

Data Interpretation

Deductive Logic Problems

Functions

Limits

Continuity