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.1Introduce A VariableTurn 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.6Introduce A VariableStrip 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.
Bars dropped among stars cut them into groups, so counting solutions is just counting bar placements.
▸ Why?
Each arrangement of bars gives exactly one list of group sizes, so the two match up without leftovers.
▸ Why?
Choosing which slots hold bars is a free selection, so the count follows the usual choosing rule.
Solve the count equation for N
We need C(N, 4)=1001. Since 1001=7 · 11 · 13, try N=14: C(14, 4)==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