Quant Tips And Tricks By Indrajeet - CEO Of iQuanta

So  ∫ 1/x = ln(x)  (Log base e )

So your summation of HP 1/1 + 1/2 + 1/3 ...+ 1/x = lnx.

You guys just remember the final result.

2) First digit of a number, yes first digit not last which everyone knows .

If we need to find the first digit of a number N , then first find the value of  {logN}, let's say equal to "m" , where {x} is the fractional part of the function. Then the first digit will be given by 10^m .

For example first digit of 3^53 :
{Log3^53} = {53log3} = 0.28
So first digit is 10^0.28 = 1.

Here is how to calculate 10^0.28.
We know 10^log2 = 2, and log2 = 0.3
So 0.28 < 0.3, hence 10^0.28 = 1.x..
Hence first digit = 1.

(Here is how to calculate 10^0.28.

We know 10^log2 = 2, and log2 = 0.3

So 0.28 < 0.3, hence 10^0.28 = 1.x..

Hence first digit = 1.)

3) Total numbers such that sum of factorial of their digits = Number itself
Only 4 such numbers are there in the number system .
a) 1!=1
b) 2!=2
c) 1!+4!+5!=145
d) 4!+0!+5!+8!+5! = 40585

4)  1*1! + 2*2! + 3*3+ ....n*n! = (n+1)! -1
Many might know this but few find hard to get it's derivation
=> (2-1)*1! + (3-1)*2! +.... (n+1- 1)n!
=> 2*1! - 1! + 3*2! - 2! + 4*3! -3! .... (n+1)n! - n!
=> 2! - 1! + 3! - 2! + 4! - 3! +. .. (n+1)! - n!
Observe the pattern . So all the terms gets cancelled except first and last term So just left with  (n+1)! - 1.

5) Number of primes between

1 - 30 => 10 primes
1 - 50 => 15 primes
1 - 100 => 25 primes
1 - 200 => 46 primes
1 - 1000 => 168 primes

Also
1001 = 13k ( multiple of 13)
1003 = 17k ( multiple of 17)
1007 = 19k ( multiple of 19)

So 1009  is the smallest 4 digit prime.

6) Number of positive integral solutions for a/x + b/y = 1/n  is the factors of a*b*n^2
And integral solutions = 2 (factors of a*b*n^2) -1

Ex : positive integral solutions of 2/x + 3/y =1/7 is factors of 2*3*7^2 => 2*2*3 = 12
And integral solutions is 2*12 - 1 = 23 .

Explanation  :
2/x + 3/y = 1/7
=> 14y + 21x - xy = 0
=>  y(14-x) + 21x = 0
=> y(14-x) + 21x -14*21 = - 14*21
=> y (14-x) - 21(14-x) = - 14*21
=> (x-14)(y-21) = 14*21= 2*3*7^2
=> a*b = 2*3*7^2,

So it's just to find in how many ways the number 2*3*7^2 can be represented as product of two numbers which is = number of its factors . Hence the formula \m/

7) Concept : A number to be written as sum of two squares :  x^2 + y^2 = N
It will have integer solns only if N contains primes of the form 4k+1 type.
( it means 11, 33, 94 etc type numbers can't be represented as sum of squares but 5^3*13, 17*37^2 can be represented )
If all primes are 4k+1 type Then positive solns => number of factors of N,

Number of positive solutions of x^2 + y^2 = 5^3*13^2 is 4*3 = 12
If it contains some 4k+3 primes with odd power then number of solutions will be 0.
If it contains even powers of 4k+3 primes, then positive solns = number of factors of 4k+1 form only.
Integral solns = 4*positive solns.

(There are a bit more complications when the number is a perfect square, which you can know by joining the I quanta course )

8) The number of ways of distributing 6 distinct rings in 4 fingers isn't 4^6.
Most people do this way including few standard  text books . But r^n is wrong here .
We apply r^n in case of balls and boxes where internal arrangement inside a box doesn't matter while in a finger the internal arrangement of ring matters and hence we can't apply r^n .
So how to solve ? First distribute 6 rings  into 4 groups then arrange .
a + b + c + d = 6, => (6+4-1)C (4-1) = 9C3
Now arrange 6 rings in 6! Ways hence your answer is 9C3*6! .

9)In the entire natural system there are only 5 such numbers which are sum of the cubes of their digits . These are known as Armstrong Numbers.

1)  1 = 1^3

2) 153 = 1^3 + 5^3 + 3^3

3) 370 = 3^3 + 7^3 + 0^3

4) 371 = 3^3 + 7^3 + 1^1

5) 407 = 4^3 + 0^3 + 7^3

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