This is continuation of my earlier article:
In this article I am going to twist the problem in part 1, to take the challenge a notch higher.
Solve for x?
Solution:
The sum of fractions (within the square brackets) cannot be zero. Hence, we have
Hence, we have
Solve for x?
This is a challenging problem from the Math Olympiad.
Many a times the hint to solve the problem lies in the problem itself. Here we have a sum of three fractions on the left side of the equation. Also, we have a 3 on the right side. This means we have a 1 for each of the fraction. Thus, we can have
Let us now shift the three 1s to the left side of the equation. Thus, we now have
The sum of fractions (within the square brackets) cannot be zero. Hence, we have
In part 2 of this article, we will twist this problem to turn the challenge a notch higher.
In most of the Indian schools and colleges we have benches which can sit two students. In my class of the engineering college in Satara we had three students who were good friends. Coincidentally all had the same last name Jadhav. The Jadhav trio used to stick together to such an extent that they used to sit together on the same bench meant for two students. Our professors used to advise them to be comfortable by sitting on separate benches. But the trio use to suffer all the discomfiture by always sharing the same bench.
Let us say that we were a batch of 40 students and were made to stand in a single row. It is not necessary to mention that the Jadhav trio will stick together even in this row. So, in how many ways can we arrange our batch of students in a row?
Since, the Jadhav trio stick together they can be considered as a unit. Hence, we now have to arrange 38 distinguishable objects. This can be done in 38! Ways.
Though the Jadhav trio stick together, they can arrange themselves in 3! = 6 ways. Hence, when our batch would stand in a row, it could arrange itself in 38! 3! Ways.
One of the most significant things that ancient India has given to the world of mathematics is the place value number system. In this system we can use any number as our base. Ten is the preferred base for the obvious reason that we have ten fingers in our two hands.
A quantity in place value number system can be represented as a polynomial in the chosen base number. This is as given below:
Refer my earlier articles:
https://chandrahasblogs.wordpress.com/2020/04/22/how-pingala-created-the-binary-number-system/
https://chandrahasblogs.wordpress.com/2020/05/18/pingalas-algorithm-for-binary-conversion/
As an example, let us convert the decimal number 235 into binary.
Since, we want to convert a decimal number 235 into binary we will divide it by 2.
Step 1) 235/2
Quotient = 117, remainder = 1
Thus, 1 will be the digit in the lowest value place.
Step 2) 117/2
Quotient = 58, remainder = 1
Step 3) 58/2
Quotient = 29, remainder = 0
Step 4) 29/2
Quotient = 14, remainder = 1
Step 5) 14/2
Quotient = 7, remainder = 0
Step 6) 7/2
Quotient = 3, remainder = 1
Step 7) 3/2
Quotient = 1, remainder = 1
Step 8) 1/2
Quotient = 0, remainder = 1
Since here, the quotient is 0, the division process comes to an end.
Thus, the binary representation of the decimal number 235 is 1110 1011.
Now, how can we convert the base of fractional quantities?
To get the digits in the lower value places go on repeating the above procedure. The process stops when the multiplicand becomes zero.
As an example, let us convert the fractional part of the decimal number 235.671875 into binary.
Since, we want to convert the fractional part of the decimal number into binary we will multiply it by 2.
Step 1) 0.671875 * 2 = 1.34375
Hence, the first digit after binary point = 1
Step 2) 0.34375 * 2 = 0.6875
Hence, the second digit after binary point = 0
Step 3) 0.6875 * 2 = 1.375
Hence, the third digit after binary point = 1
Step 4) 0.375 * 2 = 0.75
Hence, the fourth digit after binary point = 0
Step 5) 0.75 * 2 = 1.5
Hence, the fifth digit after binary point = 1
Step 6) 0. 5 * 2 = 1.0
Hence, the sixth digit after binary point = 1
Since here, the multiplicand is 0, the division process comes to an end.
Thus, the binary representation of the decimal number 235.671875 is 11101011.101011.
This article is in continuation of my earlier article:
https://chandrahasblogs.wordpress.com/2021/07/30/how-many-laddoos/
In my earlier article we had derived the formula for sum of cubes of first n natural numbers. Which is given by:
It can be seen from the formula that the sum of cubes of first n natural numbers is equal to the square of nth triangular number. In this article we shall graphically visualise this formula. For this refer figure 1.
Figure 1
If we look at the length and width of figure 1, we are adding numbers from 1 to 4 and then squaring it. Let T represent the total number of balls in figure 1. Then we have
Looking at figure 1 from a different perspective, it is a sum of cubes of numbers from 1 to 4. The red ball is cube of 1, the number of green balls are cube of 2, the yellow balls are cube of 3 and the blue balls are cubes of 4. That is
Thus, we have
Figure 2
Let us analyse this. Refer figure 2. Here, we have the square of (n – 1)th triangular number. Now, let us add n to both the length and the width. Hence, now the total number of balls will be
We have
Substituting this in (i), we have
Thus, we have
This article is an extension to my earlier article
https://chandrahasblogs.wordpress.com/2021/07/30/how-many-laddoos/
In that article I had discussed the summation of natural numbers. I am reproducing the relevant portion of that article here.
Let us calculate the sum of first n natural numbers.
Now, let us write this sum in reverse order as given below:
Let us now add both the equations, we get
Hence, the sum of natural numbers from 1 to 100 will be
We can also prove result (i) with visual aids as shown below. Refer figure 1. Here we are adding numbers from 1 to 6. Let s represent the sum. Then we have:
Figure 1
Figure 2
Now, refer figure 2. Here we have:
In the same way, we can show that
This is an extension of my earlier article:
https://chandrahasblogs.wordpress.com/2023/05/26/conditional-probability/
In this article I want to find out the probability of event B given the information that event A has occurred. Since event A has occurred all those elements which are not in A have to be removed from the sample set. Hence, the set for event A has actually become the sample set for calculating the probability for event A given event B.
The conditional probability of event B given event A is given by:
Thus, we have
From eqn(3), we have
Thus, we have
Equating (2) with (4), we have
This is called the Bayes’ theorem.
Find
Solution:
Let us first consider the expression
Now,
Thus, we have
Solution:
Find the value of
Thus, we have
And 2 is not Complex.
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