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09 Oct 2021, 03:13
Kudos
Bookmarks
Hi People,
I am starting a thread called "THE X WHYS in QUANTS" (can also be read as THE XYs of QUANTS
). What I intend to do, here, is to ask some WHYs in Quants i.e. Ask a why (or few whys) whenever I am asked to follow a formula or some formulae for something, Ask a why (or few whys) whenever I am told to apply a certain concept because that thing works in a certain way. Basically, I am just going to try and demystify some formulae and concepts in Quants.
Why I am trying to do this and why is this important :
1. Remembering formulas is just grunt work and it's boring. Probably if we can see how a particular formula is derived in first place, it could make it interesting or it could even eliminate the need to remember the formula, in certain cases.
2. Even if we were to remember formulas, couldn't it be easier to remember them well, if we also know how they were derived in first place. I think, it could.
3. Personal reasons: Being in my early 40's, I thought, remembering formulas now isn't going to be as easy as it was when I took the Indian tests CAT & XAT in my 20's. So, I started deriving formulas in my notebook, hoping to eliminate the need to remember a lot of formulas or to help myself remember the formulas well for the ones I need to.
Now, I thought I will share those with everyone to get good karma and to get enough Kudos to unlock the Gmatclub tests.
I am fully aware that this is nothing novel or new that I am trying to do here. And I am full of appreciation and gratitude towards all the mathematicians who worked on and derived these concepts and formulas and towards all the people in Gmatclub who were or are trying to help thousands of test-takers understand quant concepts. Please feel free to correct me if there are mistakes in the way I have approached something or assumed something.
--------------------
1. Why is the formula (p+1) (q+1)... for finding the no of factors of a number \(n = a^p * b^q..\)
Let's try to understand this formula with an example
Let's take \(72 = 2^3 * 3^2\)
The first term \(2^3\) has three factors (barring 1) 2, 4, 8
The second term \(3^2\) has two factors (barring 1) 3, 9
We already have 5 (3+2) factors with us. Now each of these factors can multiply with each other, creating more number of factors. The number of such factors obtained is 6 (3*2). Now, the total number of factors is 11 (5+6). "1" is divisor of all numbers and hence an additional factor to be added to 11. So, the total number of factors is 12 (3+2+6+1)
Now let's look at the formula (p+1) * (q+1) * (r+1). In this case it is (3 + 1) * (2 + 1)
Now when we multiply 3 with each term of (2 + 1), we get the 6 (3 * 2) from the above summation & the 3 (3*1), the three original factors from 2^3 and from the above summation.
When we multiply 2 (from (2+1)) with the "1" (from (3+1)), we get 2 (2*1), the two original factors from 3^2 and from the above summation
And when we multiply 1 (from (2+1)) with the "1" (from (3+1)), we get 1 (1*1) from the above summation
Thus we get the total number of factors as 12 (3+2+6+1) and hence the formula (p+1) * (q+1) * (r+1)...
--------------------
2. Why is \(\frac{(a^{p+1} - 1) * (b^{q+1} - 1) }{ (a-1) * (b-1)}\\
\) the formula for finding the sum of all factors of a number \(n = a^p * b^q\)
Let's take the same \(72 = 2^3 * 3^2\)
Now let's add all the 12 factors.
\(\\
Sum of the factors = (2^3 * 3 ^ 2) + (2^3 * 3^1) + (2^3 * 3^0) + (2^2 * 3^2) + (2^2 * 3^1) + (2^2 * 3^0) + (2^1 * 3^2) + (2^1 * 3^1) + (2^1 * 3^0) + ( 2^0 * 3^2) + (2^0 * 3^1) + (2^0 * 3^0)\)
\( = 2^3 * (3^2 + 3 + 1) + 2^2 * (3^2 + 3 + 1) + 2 * ( 3^2 + 3 + 1) + 1 * ( 3^2 + 3 + 1)\)
\( = (2^3 + 2^2 + 2 + 1) * (3^2 + 3 + 1)\)
Now we know that \((a^n - b^n) = (a - b ) * ( a^{n-1} + a^{n-2} * b + a^{n-3} * b^2 + ......+ b^{n-1})\)
Now compare \((2^3 + 2^2 + 2 + 1)\) with \(( a^{n-1} + a^{n-2} * b + a^{n-3} * b^2 + ......+ b^{n-1})\)
\((2^3 + 2^2 + 2 + 1)\) is nothing but \(\frac{(2^4 - 1^4) }{ (2-1)}\)
And \(\frac{(2^4 - 1^4) }{ (2-1)}\) is nothing but \(\frac{( 2 ^ {3+1} - 1 ) }{ (2 - 1)}\) i.e. \(\frac{(a^{p+1} - 1) }{ (a-1)}\)
So, when we apply the above for all prime factors with any power, we get \(\frac{(a^{p+1} - 1) }{ (a-1)}\) * \(\frac{(b^{q+1} - 1) }{ (b-1)}\)
-------------------------
To be continued tomorrow.... . Time to study now 
I am starting a thread called "THE X WHYS in QUANTS" (can also be read as THE XYs of QUANTS
). What I intend to do, here, is to ask some WHYs in Quants i.e. Ask a why (or few whys) whenever I am asked to follow a formula or some formulae for something, Ask a why (or few whys) whenever I am told to apply a certain concept because that thing works in a certain way. Basically, I am just going to try and demystify some formulae and concepts in Quants. Why I am trying to do this and why is this important :
1. Remembering formulas is just grunt work and it's boring. Probably if we can see how a particular formula is derived in first place, it could make it interesting or it could even eliminate the need to remember the formula, in certain cases.
2. Even if we were to remember formulas, couldn't it be easier to remember them well, if we also know how they were derived in first place. I think, it could.
3. Personal reasons: Being in my early 40's, I thought, remembering formulas now isn't going to be as easy as it was when I took the Indian tests CAT & XAT in my 20's. So, I started deriving formulas in my notebook, hoping to eliminate the need to remember a lot of formulas or to help myself remember the formulas well for the ones I need to.
Now, I thought I will share those with everyone to get good karma and to get enough Kudos to unlock the Gmatclub tests.
I am fully aware that this is nothing novel or new that I am trying to do here. And I am full of appreciation and gratitude towards all the mathematicians who worked on and derived these concepts and formulas and towards all the people in Gmatclub who were or are trying to help thousands of test-takers understand quant concepts. Please feel free to correct me if there are mistakes in the way I have approached something or assumed something.
--------------------
1. Why is the formula (p+1) (q+1)... for finding the no of factors of a number \(n = a^p * b^q..\)
Let's try to understand this formula with an example
Let's take \(72 = 2^3 * 3^2\)
The first term \(2^3\) has three factors (barring 1) 2, 4, 8
The second term \(3^2\) has two factors (barring 1) 3, 9
We already have 5 (3+2) factors with us. Now each of these factors can multiply with each other, creating more number of factors. The number of such factors obtained is 6 (3*2). Now, the total number of factors is 11 (5+6). "1" is divisor of all numbers and hence an additional factor to be added to 11. So, the total number of factors is 12 (3+2+6+1)
Now let's look at the formula (p+1) * (q+1) * (r+1). In this case it is (3 + 1) * (2 + 1)
Now when we multiply 3 with each term of (2 + 1), we get the 6 (3 * 2) from the above summation & the 3 (3*1), the three original factors from 2^3 and from the above summation.
When we multiply 2 (from (2+1)) with the "1" (from (3+1)), we get 2 (2*1), the two original factors from 3^2 and from the above summation
And when we multiply 1 (from (2+1)) with the "1" (from (3+1)), we get 1 (1*1) from the above summation
Thus we get the total number of factors as 12 (3+2+6+1) and hence the formula (p+1) * (q+1) * (r+1)...
--------------------
2. Why is \(\frac{(a^{p+1} - 1) * (b^{q+1} - 1) }{ (a-1) * (b-1)}\\
\) the formula for finding the sum of all factors of a number \(n = a^p * b^q\)
Let's take the same \(72 = 2^3 * 3^2\)
Now let's add all the 12 factors.
\(\\
Sum of the factors = (2^3 * 3 ^ 2) + (2^3 * 3^1) + (2^3 * 3^0) + (2^2 * 3^2) + (2^2 * 3^1) + (2^2 * 3^0) + (2^1 * 3^2) + (2^1 * 3^1) + (2^1 * 3^0) + ( 2^0 * 3^2) + (2^0 * 3^1) + (2^0 * 3^0)\)
\( = 2^3 * (3^2 + 3 + 1) + 2^2 * (3^2 + 3 + 1) + 2 * ( 3^2 + 3 + 1) + 1 * ( 3^2 + 3 + 1)\)
\( = (2^3 + 2^2 + 2 + 1) * (3^2 + 3 + 1)\)
Now we know that \((a^n - b^n) = (a - b ) * ( a^{n-1} + a^{n-2} * b + a^{n-3} * b^2 + ......+ b^{n-1})\)
Now compare \((2^3 + 2^2 + 2 + 1)\) with \(( a^{n-1} + a^{n-2} * b + a^{n-3} * b^2 + ......+ b^{n-1})\)
\((2^3 + 2^2 + 2 + 1)\) is nothing but \(\frac{(2^4 - 1^4) }{ (2-1)}\)
And \(\frac{(2^4 - 1^4) }{ (2-1)}\) is nothing but \(\frac{( 2 ^ {3+1} - 1 ) }{ (2 - 1)}\) i.e. \(\frac{(a^{p+1} - 1) }{ (a-1)}\)
So, when we apply the above for all prime factors with any power, we get \(\frac{(a^{p+1} - 1) }{ (a-1)}\) * \(\frac{(b^{q+1} - 1) }{ (b-1)}\)
-------------------------
To be continued tomorrow....
. Time to study now
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12 Nov 2022, 04:15
Kudos
Bookmarks
This is link for the first post - https://gmatclub.com/forum/the-x-whys-1-why-2-whys-5-whys-etc-thread-in-quants-371079.html (the first post also talks about what's this thread all about)
This is the second post in the series. I know that I am continuing the series after a year. I started it with all enthusiasm but then I quickly realized that doing this series might take a lot of time away from studying; hence, I didn't continue the series. I thought of giving this a try again, and that's why doing this post now.
The formula for finding the sum of first N positive odd numbers is \(N^2\). Why ?
Let's derive this formula with an example '1 + 3 + 5 + 7' , the sum of first four positive odd numbers.
1 + 3 + 5 + 7
= 1 + (1 + 1*2) + (1 + 2*2) + (1 + 2*3)
= 4 + (1*2) + (2*2) + (3*2)
= 4 + 2*(1+2+3)
N = 4, so 4 + 2*(1+2+3) is nothing but
N + 2 * ((N-1)*N/2)
(the sum of first X natural numbers is X*(X+1)/2. Here we want the sum of 1 + 2 + 3 i.e. sum of (N-1) numbers.
By replacing X with (N-1) , we get (N-1)*N/2.)
Cancelling 2, we get N + (N-1)*N
= N + \(N^2\) - N
= \(N^2\)
This is the second post in the series. I know that I am continuing the series after a year. I started it with all enthusiasm but then I quickly realized that doing this series might take a lot of time away from studying; hence, I didn't continue the series. I thought of giving this a try again, and that's why doing this post now.
The formula for finding the sum of first N positive odd numbers is \(N^2\). Why ?
Let's derive this formula with an example '1 + 3 + 5 + 7' , the sum of first four positive odd numbers.
1 + 3 + 5 + 7
= 1 + (1 + 1*2) + (1 + 2*2) + (1 + 2*3)
= 4 + (1*2) + (2*2) + (3*2)
= 4 + 2*(1+2+3)
N = 4, so 4 + 2*(1+2+3) is nothing but
N + 2 * ((N-1)*N/2)
(the sum of first X natural numbers is X*(X+1)/2. Here we want the sum of 1 + 2 + 3 i.e. sum of (N-1) numbers.
By replacing X with (N-1) , we get (N-1)*N/2.)
Cancelling 2, we get N + (N-1)*N
= N + \(N^2\) - N
= \(N^2\)
12 Nov 2022, 04:18
Kudos
Bookmarks
Here's the link for my second post in the series - https://gmatclub.com/forum/the-x-whys-in-quants-2nd-post-why-n-2-sum-of-1st-ve-odd-nos-401451.html
The formula for finding the sum of first N positive odd numbers is N^2. Why ?
The formula for finding the sum of first N positive odd numbers is N^2. Why ?
