Quant Tips And Tricks By Indrajeet – CEO Of iQuanta  Indrajeet Singh, the CEO of iQuanta gave a new direction to the world of online teaching where the doubts gets cleared within seconds or minutes using the previously neglected medium, Facebook as his platform. He started iQuanta last year, and it delivered splendid results in its very first year with two 100%ilers, sixteen 99.6+ %ilers & around 30 final converts from IIM A,B,C and XLRI. iQuanta is mostly about unconventional methods and concepts which aren’t taught in conventional coaching centres and his free peer learning platform i.e. CAT Preparation group is 93k members strong and is highly active – https://www.facebook.com/groups/Rockthecat/

Here are few of his techniques:

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.

(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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If you have any queries, you can get it clarified by posting in the ‘Comments Section’ below.

Indrajeet iQuanta

Indrajeet Singh, the CEO of iQuanta gave a new direction to the world of online teaching where the doubts gets cleared within seconds or minutes using the previously neglected medium, Facebook as his platform. He started iQuanta last year, and it delivered splendid results in its very first year with two 100%ilers, sixteen 99.6+ %ilers & around 30 final converts from IIM A,B,C and XLRI. iQuanta is mostly about unconventional methods and concepts which aren’t taught in conventional coaching centres and his free peer learning platform i.e. CAT Preparation group is 93k members strong and is highly active – https://www.facebook.com/groups/Rockthecat/   