What is Subtracted from 1000 to Get a Perfect Square?

Mathematics often involves intriguing problems that challenge our understanding of numbers and their properties. One such problem is finding the values that can be subtracted from 1000 to obtain a perfect square. This article will explore this concept in detail, providing a clear explanation and examples to make it accessible to all readers.

Defining the Problem

Given a number 1000, we seek a value ( x ) such that when subtracted from 1000, the result is a perfect square. Mathematically, we express this as:

[1000 - x n^2]

where ( n ) is an integer. Rearranging the equation, we get:

[x 1000 - n^2]

Determining the Range for ( n )

To ensure ( x ) remains a non-negative number, we require:

[1000 - n^2 geq 0]

Solving for ( n ), we take the square root of both sides:

[n leq sqrt{1000} approx 31.62]

Hence, ( n ) can be any integer from 0 to 31.

Calculating ( x ) for Each ( n )

We will now calculate ( x ) for each integer value of ( n ) from 0 to 31:

For ( n 0 ): ( x 1000 - 0^2 1000 ) For ( n 1 ): ( x 1000 - 1^2 999 ) For ( n 2 ): ( x 1000 - 2^2 996 ) For ( n 3 ): ( x 1000 - 3^2 991 ) For ( n 4 ): ( x 1000 - 4^2 984 ) For ( n 5 ): ( x 1000 - 5^2 975 ) For ( n 6 ): ( x 1000 - 6^2 964 ) For ( n 7 ): ( x 1000 - 7^2 951 ) For ( n 8 ): ( x 1000 - 8^2 936 ) For ( n 9 ): ( x 1000 - 9^2 919 ) For ( n 10 ): ( x 1000 - 10^2 900 ) For ( n 11 ): ( x 1000 - 11^2 879 ) For ( n 12 ): ( x 1000 - 12^2 856 ) For ( n 13 ): ( x 1000 - 13^2 831 ) For ( n 14 ): ( x 1000 - 14^2 804 ) For ( n 15 ): ( x 1000 - 15^2 775 ) For ( n 16 ): ( x 1000 - 16^2 736 ) For ( n 17 ): ( x 1000 - 17^2 731 ) For ( n 18 ): ( x 1000 - 18^2 676 ) For ( n 19 ): ( x 1000 - 19^2 661 ) For ( n 20 ): ( x 1000 - 20^2 600 ) For ( n 21 ): ( x 1000 - 21^2 579 ) For ( n 22 ): ( x 1000 - 22^2 524 ) For ( n 23 ): ( x 1000 - 23^2 531 ) For ( n 24 ): ( x 1000 - 24^2 460 ) For ( n 25 ): ( x 1000 - 25^2 375 ) For ( n 26 ): ( x 1000 - 26^2 276 ) For ( n 27 ): ( x 1000 - 27^2 229 ) For ( n 28 ): ( x 1000 - 28^2 184 ) For ( n 29 ): ( x 1000 - 29^2 141 ) For ( n 30 ): ( x 1000 - 30^2 100 ) For ( n 31 ): ( x 1000 - 31^2 39 )

Examples of Perfect Squares

From these calculations, we see that various values can be subtracted from 1000 to obtain perfect squares. For example:

Subtracting 100 gives ( 900 30^2 ) Subtracting 39 gives ( 961 31^2 )

Conclusion

In conclusion, several integers can be subtracted from 1000 to yield perfect squares. This article has provided a detailed explanation and examples, demonstrating the ease with which we can solve such problems using basic arithmetic and algebraic manipulation. By understanding the relationship between numbers and perfect squares, you can explore similar problems and solutions with confidence and precision.