Mathemagics by Indrajeet

1) Summation of the series 1/1 + 1/2 + 1/3 .... + 1/x = ?
It's summation is kind of difficult to find and not known to many . So engineers can just integrate 1/x from 1 to x
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 = 1.3) 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
= > (21)*1! + (31)*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(14x) + 21x = 0
= > y(14x) + 21x 14*21 =  14*21
= > y (14x)  21(14x) =  14*21
= > (x14)(y21) = 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 like emoticon
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 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+41)C (41) = 9C3
Now arrange 6 rings in 6! Ways hence your answer is 9C3*6! .