Solving for Consecutive Odd Numbers Using Algebra and J Programming

In this article, we will explore how to solve a problem involving consecutive odd numbers using both algebraic methods and the J programming language. The problem is as follows:

There are 5 consecutive odd numbers. If the difference between the square of the average of the first two odd numbers and the square of the average of the last two odd numbers is 492, what is the smallest number?

Algebraic Method

Let's denote the five consecutive odd numbers as n, n 2, n 4, n 6, n 8.

Step 1: Find the averages of the first two and the last two odd numbers.

Average of the first two odd numbers:

[text{Average}_1 frac{n (n 2)}{2} n 1]

Average of the last two odd numbers:

[text{Average}_2 frac{(n 6) (n 8)}{2} n 7]

Step 2: Calculate the squares of the averages.

Square of the average of the first two odd numbers:

[text{Square}_1 (n 1)^2 n^2 2n 1]

Square of the average of the last two odd numbers:

[text{Square}_2 (n 7)^2 n^2 14n 49]

Step 3: Find the difference between the squares.

We need to calculate the difference between these squares:

[text{Difference} text{Square}_1 - text{Square}_2] [text{Difference} (n^2 2n 1) - (n^2 14n 49) -12n - 48]

Step 4: Set the difference equal to 492.

According to the problem, this difference equals 492:

[-12n - 48 492]

Step 5: Solve for n.

First, add 48 to both sides:

[-12n 540]

Now, divide by -12:

[n frac{540}{-12} -45]

Therefore, the smallest number is -45. The five consecutive odd numbers are:

[-45, -43, -41, -39, -37]

So, the smallest of these five consecutive odd numbers is -45.

Brute Force Solution Using J Programming Language

The J programming language can be used to solve the same problem using a brute force approach.

The task is to find the five consecutive odd integers that meet the criteria:

: ea avg ea ({3 4{3 4 5 6 7 8 9 10})37 39 41 43 45

The smallest odd number is 37.

Another Algebraic Approach

Another algebraic approach, defined as the middle number x, can be implemented as follows:

Numbers: x-4, x-2, x, x 2, x 4 Average of first two: x-3 Average of last two: x 3 Equation: (x 3)^2 - (x-3)^2 492

Transform the equation:

[(x 3)^2 - (x-3)^2 492] [(x^2 6x 9) - (x^2 - 6x 9) 492] [12x 492] [x 41]

The smallest number in this case is x-4, which is 37.

Yet Another Algebraic Method

The numbers are a, b, c, d, e

a^2 b b^2 c a^4 c^2 d a^6 d^2 e a^8 (e * d) / 2^2 - (a * b) / 2^2 492

Putting the equation in terms of a:

[(a^8 * a^6) / 2^2 - (a^2 * a^4) / 2^2 492] [a^{14} - a^6 492]

This can be simplified to:

[a^2 * a^4 - a^2 492] [a^2 * (a^4 - 1) 492]

Solving for a in the simplified form:

[a^2 37]

Checking the work:

The numbers are 37, 39, 41, 43, 45 Average of first two is 38, square of 38 is 1444 Average of last two is 44, square of 44 is 1936 1936 - 1444 492

It appears to be correct.