Category Archives: Mathematics

Math Update Sommer Semester

https://www.mathsisfun.com/calculus/implicit-differentiation.html

https://www.mathway.com/popular-problems/Algebra/200043

 

Mental Math Tricks

Will share that soon

Area of Circle using function

for formula: http://www.mathsisfun.com/area.htm

code:

We have to always remember that value that define is always pass from main function to customized function named by user so here r=3 is passing to function britto and then the britto formula pi*r*r is applying and returned the value and it is printed in the main function where the variable defined as khetrofol.

Here pi declared as global variable with it’s value.

Find Maximum and Minimum and Summation value from array

Did this problem from subeen vai’s book.

For maximum value:

For minimum value:

Summation:

Average:

Will give here

 

 

Math Motivation

http://www.prothom-alo.com/bangladesh/article/580627/%E0%A6%97%E0%A6%A3%E0%A6%BF%E0%A6%A4-%E0%A6%85%E0%A6%B2%E0%A6%BF%E0%A6%AE%E0%A7%8D%E0%A6%AA%E0%A6%BF%E0%A7%9F%E0%A6%BE%E0%A6%A1%E0%A7%87%E0%A6%B0-%E0%A6%A4%E0%A6%BE%E0%A6%B0%E0%A6%95%E0%A6%BE%E0%A6%B0%E0%A6%BE-%E0%A6%95%E0%A7%87-%E0%A6%95%E0%A7%8B%E0%A6%A5%E0%A6%BE%E0%A7%9F

LCM – Least Common Multiple

Multiple of 3 is
3×1=3
3×2=6
3×3=9
3×4=12
3×5=15
3×6=18
3×7=21
3×8=24
3×9=27
3×10=30

Multiple of 4 is
4×1=4
4×2=8
4×3=12
4×4=16
4×5=20
4×6=24
4×7=28
4×8=32
4×9=36
4×10=40

Here in 3 and 4 mutilples common is 12
and it is only one value and so it is the lowest also then but if there were some more values than result could be something changed.

So here LCM=12
Here I implemented the code
applied this formula
lcm(a,b)=a*b/gcd(a,b);

code goes here :

 

GCD Greatest Common Divisor(Factor)/Highest Common Factor

এখানে লজিকটা হচ্ছে…Here the logic is:
For 12:
1×12=12
2×6=12
3×4=12
So factor of 12=1,2,3,4,6,12
(We don’t need the negative values here)

For 30:
1×30=30
2×15=30
3×10=30
5×6=30
So factor of 30=1,2,3,5,6,10,15,30

Now which is the common factor among 12 and 30
So factor of 12=1,2,3,4,6,12
So factor of 30=1,2,3,5,6,10,15,30

common factor=1,2,3,6
Largest common factor here=6

So now if we divide 12/6 then we get 2
and for 30/6 we get 5

that is all divided perfectly so the logic should be like this

1.Find all the factors of each number
2. Circle the common factors
3. Choose the greatest of those
4. Divide the existing number with greatest common factor thern also get a perfect divide

Here is my code:

gcd

Here is a link that described many thinks about this clearly: http://www.mathsisfun.com/greatest-common-factor.html

 

Another method can also apply here:
link for understanding the modarithmetic euclidean algorithmhttps://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/the-euclidean-algorithm

Code: