AMC 10 · 2010 · #21
Grade 7 algebraPick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #14 (Extreme Principle): the question asks for the smallest a, so once we know what a depends on we push that quantity to its minimum. Tool #4 (Introduce a Variable): name the three roots p,q,r so the polynomial becomes a product of linear factors we can read coefficients from. Tool #7 (Identify Subproblems): split the work into (i) link a to the roots, (ii) factor 2010, (iii) group the factors, (iv) minimize. Tool #2 (Make a Systematic List): there are only a few ways to split 2010 into three whole-number factors, so list them all and compare their sums.
Name the roots and factor the cubic
Name the three positive integer roots p, q, r; then the same cubic can be rewritten as the product (x-p)(x-q)(x-r).
A polynomial's roots are exactly the numbers that make each factor zero, so writing it as a product of (x-root) pieces puts the roots in plain sight.
6.EE.A.2Introduce A VariableMatch coefficients to pin down a
Expanding and matching terms, the constant gives -pqr and the x² term gives -(p+q+r), so pqr=2010 and a=p+q+r.
Expanding the factored form and comparing like terms turns the hidden coefficient a into something concrete: the sum of the roots.
Expanding the factored form and comparing like terms turns the hidden coefficient into the sum of the roots.
▸ Why?
Two expressions that agree for every input must match term by term.
▸ Why?
That match is exactly what turns the coefficients into the sum and product of the roots.
Break 2010 into its prime factors
Dividing step by step gives 2010=2·3·5·67 — four primes for only three roots, so two of them must merge.
Every whole number is a fixed bag of primes, so any set of factors of 2010 is just a way of dividing that bag {2,3,5,67} into groups.
4.OA.B.4Identify SubproblemsList every way to make three factors
There are six pairs to merge, hence six triples, and their sums are 78, 80, 84, 142, 208, 340.
Listing all the groupings guarantees we do not miss a smaller sum hiding among the possibilities.
4.OA.B.4Make A Systematic ListPick the smallest sum
Merging the two smallest primes 2 and 3 leaves roots 6, 5, 67, the lowest total, so a is 78 — choice (A).
With a fixed product, a sum of factors shrinks when the factors are kept as balanced as possible, so merging the two smallest primes wastes the least.
6.EE.B.5Extreme PrincipleThe three roots multiply to 2010 and a is their sum, so split 2010's primes 2·3·5·67 into three factors as balanced as possible: 6, 5, 67 give a=78.
- Name the roots and factor the cubic
- Match coefficients to pin down a
- Break 2010 into its prime factors
- List every way to make three factors
- Pick the smallest sum