AMC 10 · 2016 · #20
Grade 7 algebraPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Each combined term is built by choosing one of the five pieces a,b,c,d,1 from each of the N factors, so a term is fully described by how many factors contributed each piece. Tool #4 (Introduce a Variable) names those five counts and turns 'count the terms' into 'count whole-number solutions of i+j+k+l+m=N.' Tool #9 (Solve an Easier Related Problem) removes the awkward 'each variable at least once' rule by subtracting one from each variable's count, leaving a clean all-nonnegative version. Tool #2 (Make a Systematic List), in its stars-and-bars form, counts those solutions in one shot as C(N, 4). Finally tool #6 (Guess and Check) finds the N with C(N, 4)=1001, helped by noticing 1001=7· 11· 13.
Describe one qualifying term
Each factor donates one of a, b, c, d, 1, so a term is a^ib^jc^kd^l1^m where the five exponent counts add up to N.
Each factor hands over exactly one letter or a 1, so a term is just a tally of how many factors gave each piece.
6.EE.A.1Use Matrix LogicTurn the condition into an equation
A term qualifies only when i, j, k, l are each at least 1 (m may be 0), so the count equals the number of solutions of i+j+k+l+m=N.
Counting terms is the same as counting the exponent lists allowed by the rules.
6.EE.B.6Use Matrix LogicStrip out the 'at least one' rule
Give each of a, b, c, d its required 1 up front via i'=i-1, etc.; every variable is now 0 or more and the total drops to N-4.
Hand each variable its mandatory 1 first, and what's left is a no-strings counting problem.
6.EE.B.7Solve An Easier Related ProblemCount the solutions with stars and bars
With N-4 stars and 4 bars in a row, the bars split the stars into 5 groups; choosing which 4 of the N positions are bars gives C(N, 4).
Bars dropped among stars cut them into groups, so counting solutions is just counting bar placements.
7.SP.C.8Make A Systematic ListSolve the count equation for N
We need C(N, 4)=1001. Since 1001=7· 11· 13, try N=14: C(14, 4)=(14· 13· 12· 11)/24=1001, and C(N, 4) only grows, so N=14.
Seeing 1001=7· 11· 13 points straight at 14· 13· 12· 11 over 24.
7.NS.A.3Guess And CheckEach term is just a tally of how many factors give each letter, so counting the terms turns into a stars-and-bars count of C(N, 4), and C(14, 4)=1001 gives N=14.
- Describe one qualifying term
- Turn the condition into an equation
- Strip out the 'at least one' rule
- Count the solutions with stars and bars
- Solve the count equation for N