1999 AHSME Problems/Problem 28
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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) | ||
| Preceded by Problem 27 | Followed by Problem 29 | |
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