2010 AMC 12B Problems/Problem 21
Contents
Problem
Let , and let be a polynomial with integer coefficients such that
,
and
.
What is the smallest possible value of ?
Solution
We observe that because , if we define a new polynomial such that , has roots when ; namely, when .
Thus since has roots when , we can factor the product out of to obtain a new polynomial such that .
Then, plugging in values of we get
Thus, the least value of must be the . Solving, we receive , so our answer is .
To complete the solution, we can let , and then try to find . We know from the above calculation that , and . Then we can let , getting . Let , then . Therefore, it is possible to choose , so the goal is accomplished. As a reference, the polynomial we get is
Solution 2
The evenly-spaced data suggests using discrete derivatives to tackle this problem. First, note that any polynomial of degree
can also be written as
Moreover, the coefficients are integers for iff the coefficients are integers for . This latter form is convenient for calculating discrete derivatives of .
The discrete derivative of a function is the related function defined as
With this definition, it's easy to see that for any positive integer we have
This in turn allows us to use successive discrete derivatives evaluated at to calculate all of the coefficients using
We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:
Thus we can read down the column for to find that for . Interestingly, even if we choose to have degree greater than , the coefficients of lowest order always satisfy these conditions. Moreover, it's straightforward to show that the of degree with satisfying these conditions will fit the data given in the problem statement. Thus, to find minimal necessary and sufficient conditions on the value of , we need only consider these equations. As a result, with integer coefficients fitting the given data exists iff divides for . In other words, it's necessary and sufficient that
,
,
,
,
,
,
, and
.
The last condition holds if divides evenly into . Since such will also satisfy the first conditions, our answer is .
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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