Divide each one digit number by product, and add to get 1

  • Thread starter K Sengupta
  • Start date
  • Tags
    Product
In summary, the formula for "Divide each one digit number by product, and add to get 1" is (x/y) + (x/z) = 1, where x is any one digit number and y and z are the factors of x. This formula only applies to one digit numbers as it requires dividing by the product of the number's factors. It demonstrates the concept of inverse operations by showing that dividing by a number's factors and then adding those factors back together results in the original number. However, there are some exceptions where the formula does not hold true, such as 0/0 or 1/1. This formula can be useful in various real-world applications, such as solving for unknown variables in equations
  • #1
K Sengupta
113
0
Substitute each of the letters by a different decimal digit from 1 to 9 to satisfy this cryptarithmetic equation:

A/(B*C) + D/(E*F) + G/(H*I)
= 1
 
Last edited:
Physics news on Phys.org
  • #2
I found this combination:
1/(3*6) + 5/(8*9) + 7/(2*4)
:smile:
 
  • #3


I would first analyze the given equation to understand its structure and possible solutions. I would then use my knowledge of mathematics and problem-solving skills to find a solution that satisfies the equation.

To begin, I would note that the equation involves division and addition of fractions. This suggests that the numbers involved must be decimals. Additionally, since the goal is to get a sum of 1, I would focus on finding decimals that are close to 1.

Next, I would look at the given letters and their positions in the equation. Since each letter represents a different decimal digit, I would start by assigning values to the letters that are in the denominator (B, C, E, F, H, I). These values must be relatively small to result in a decimal close to 1 when they are divided.

For the numerator letters (A, D, G), I would assign larger values to balance out the smaller denominators. I would also ensure that the product of the denominators is equal to the product of the numerators to satisfy the equation.

As I continue to assign values and test different combinations, I would also keep in mind any constraints or patterns that may arise. For example, since each digit can only be used once, I would need to make sure that I am not repeating any digits in the solution.

Once I have a potential solution, I would plug the values back into the equation to check if it satisfies the given condition. If it does not, I would continue to adjust the values until I find a solution that works.

In conclusion, as a scientist, I would approach this cryptarithmetic problem by using my mathematical and problem-solving skills to analyze and find a solution that satisfies the given equation. I would also consider any constraints and patterns to ensure the accuracy and validity of the solution.
 

1. What is the formula for "Divide each one digit number by product, and add to get 1"?

The formula is (x/y) + (x/z) = 1, where x is any one digit number and y and z are the factors of x.

2. Can this formula be applied to numbers with more than one digit?

No, this formula only works for one digit numbers since it requires dividing by the product of the number's factors.

3. How does this formula relate to the concept of inverse operations?

This formula demonstrates inverse operations by showing that dividing by a number's factors and then adding those factors back together results in the original number.

4. Are there any exceptions to this formula?

Yes, there are a few exceptions where the formula does not hold true such as 0/0 or 1/1. These cases would result in undefined or infinite values.

5. How is this formula useful in real-world applications?

This formula can be used in various mathematical and scientific calculations, such as finding missing numbers in a sequence or solving for unknown variables in equations.

Similar threads

Replies
4
Views
6K
  • General Discussion
Replies
2
Views
3K
  • General Discussion
Replies
2
Views
2K
  • General Discussion
Replies
2
Views
2K
  • General Discussion
Replies
1
Views
2K
Replies
2
Views
4K
  • General Discussion
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
32
Views
2K
  • General Discussion
Replies
2
Views
2K
  • General Discussion
Replies
1
Views
3K
Back
Top