Problem 3 (The Number Problem)

Among the natural numbers 1, 2, 3, .... , find all solutions (if any) of the equation x² = 1.5 y³, where both squares x² and y² are restricted to be less than 10,000.

Solution

Since both x² and y² are less than 10,000 we know x is one of the integers 1,2, ... , 99, and since 1.5 y³ = x² < 10,000, we know that y is one of the integers 1, 2, ... , 18. We could substitute all the combinations of these two numbers into the equation to see if a solution exists, but let's do some basic analysis of numbers so we don't have as many possibilities.

We argue as follows (watch carefully):

• y³ must be an even integer (or else 1.5 y³ is not an integer)
• hence x² is a multiple of 3 (since x² = 1.5 y³ )
• hence x² is a multiple of 9 (it is the square of a number)
• hence 1.5 y³ is a multiple of 9 (since 1.5 y³ = x² )
• hence y³ is a multiple of 6
• hence y is a multiple of 6; that is y = 6, 12, or 18.

Since we know that y is either 6, 12, or 18, all we have to do is form the following table:

y	y3	1.5 y3
6	216	324
12	1728	2592
18	5832	8749

and ask whether any of the numbers 1.5 y³ are perfect squares. The answer is that 324 = 18² is a perfect square, giving us the single solution of x = 13, y = 6. you can check these values by noting that 18² = 1.5 (18)².

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Last modified on Friday, February 12, 1999