Math Problem Statement

  1. The first digit of a four-digit natural number is 7. If this digit is transferred to the last place, a four-digit natural number will be obtained that is 864 less than the original number. Find out the original

Solution

Let the four-digit number be NN. Since the first digit is 7, we can express the number as:

N=7000+x,N = 7000 + x,

where xx is the number formed by the last three digits.

When the first digit 7 is moved to the end, the new number becomes:

M=10x+7.M = 10x + 7.

According to the problem, this new number is 864 less than the original number. Thus, we can write:

M=N864.M = N - 864.

Substituting the expressions for MM and NN:

10x+7=(7000+x)864.10x + 7 = (7000 + x) - 864.

Simplify the equation:

10x+7=6136+x.10x + 7 = 6136 + x.

Rearranging terms:

10xx=61367,10x - x = 6136 - 7,

9x=6129.9x = 6129.

Solving for xx:

x=61299=681.x = \frac{6129}{9} = 681.

Thus, the original number is:

N=7000+x=7000+681=7681.N = 7000 + x = 7000 + 681 = 7681.

Final Answer:

The original number is 7681.

Would you like a detailed breakdown or clarification?
Here are 5 related questions for further exploration:

  1. How can we check the correctness of the solution?
  2. What happens if the first digit was a number other than 7?
  3. Can this logic be applied to numbers with more or fewer digits?
  4. What if the number obtained after transferring the digit was a specific multiple of the original?
  5. How would the solution change if the difference was larger than 864?

Tip: Always verify the final solution by plugging it back into the original problem conditions.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations

Formulas

Representation of numbers in terms of place values
Basic linear equation solving

Theorems

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Suitable Grade Level

Grades 8-10