1999 AHSME Problems/Problem 28
Contents
Problem
Let be a sequence of integers such that (i) for (ii) ; and (iii) . Let and be the minimal and maximal possible values of , respectively. Then
Solution
Clearly, we can ignore the possibility that some are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider .
Also, order of the does not matter, so we are only interested in the counts of the variables of each type. Let of the be equal to , equal to , and equal to .
The conditions (ii) and (iii) simplify to:
(ii)
(iii)
and we want to find the maximum and minimum of over all non-negative solutions of the above two equations.
Subtracting twice (ii) from (iii) we get . By entering that into one of the two equations and simplifying we get .
Thus all the solutions of our system of equations have the form .
As all three variables must be non-negative integers, we have and .
For of the form the expression we are maximizing/minimizing simplifies to . Clearly, the maximum is achieved for and the minimum for . Their values are and , and therefore .
Note
The minimum is achieved for The maximum is achieved for
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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