Competition · AMC preparation · step 4 of 4
AMC 8 · 2012 · #13
Grade 6 number-theoryPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The price per pencil c has to divide both 143 and 187, and the problem says c ＞ 1. With Tool #13 (Convert to Algebra) we write the two simple equations j · c = 143 and s · c = 187, which makes "c is a common divisor" obvious. Then Tool #6 (Guess and Check) is the lightest way to find c — try small primes (2, 3, 5, 7, 11, …) on 143 until one fits, and confirm it also divides 187. Tool #7 (Identify Subproblems) splits the work into three clean steps: find c, then find j and s, then subtract.
Convert to cents and name the unknowns
Turn both totals into cents and name the unknowns c, j, s, so the equations are j·c = 143 and s·c = 187 with c ＞ 1.
Naming the unknown with a letter and writing the two "count × price = total" equations is the Grade 6 "use variables to represent numbers" move.
6.EE.B.6Convert To AlgebraFactor 143 to find its primes
Since c divides both totals, test small primes on 143 and find 143 = 11 × 13.
Testing 2, 3, 5, 7, 11, … until one works is the standard Grade 4 "find factor pairs" routine.
The price of one pencil in cents must divide both totals, 143 and 187, evenly — it is a common divisor of the two amounts.
▸ Why?
The price has to divide 143 with nothing left over, because Jamar's 143 cents is his whole number of pencils, each costing that one price.
▸ Why?
A total is the number of pencils multiplied by the price, so 143 is exactly that many equal groups of the price.
▸ Why?
Splitting 143 back into groups of the price just reverses the multiplication that built it, so the division leaves no remainder.
▸ Why?
The same price must also divide 187 with nothing left over, because Sharona's 187 cents is her whole number of pencils at that identical price.
▸ Why?
Her total is again the pencil count multiplied by the same price, so 187 is that many equal groups of the price.
▸ Why?
Dividing 187 by the price undoes the multiplication that produced it, so it too comes out even.
Check that 11 divides 187
Check the same prime on the other total: 187 = 11 × 17, so 11 divides both.
Confirming a common factor of two numbers is exactly the Grade 6 "greatest common factor" idea.
6.NS.B.4Guess And CheckRule out the other common divisors
The only common divisors of 143 and 187 are 1 and 11, and c ＞ 1 rules out 1, so c = 11 cents is forced.
Once we have full factor lists, comparing them is a checking step, not a leap of faith.
6.NS.B.4Guess And CheckFind each kid's pencils and subtract
Divide back: j = 13 and s = 17, so Sharona bought s - j = 4 more pencils — (C).
Dividing a 3-digit number by an 11 is a Grade 4 multi-digit division, and the subtraction is the last subproblem of the split.
4.NBT.B.6Identify SubproblemsThis AMC 8 problem just needs Grade 6 GCF thinking: find the only price bigger than 1 cent that divides both totals, then divide to get the counts.
- Convert to cents and name the unknowns
- Factor 143 to find its primes
- Check that 11 divides 187
- Rule out the other common divisors
- Find each kid's pencils and subtract
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