Competition · AMC preparation · step 4 of 4
AMC 8 · 2016 · #11
Grade 6 algebranumber-theoryPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The condition is a statement about digits, so Tool #5 (Use Variables) lets us name the tens digit t and the units digit u and write N = 10t + u in place-value form. Adding N to its reversal 10u + t collapses neatly to 11(t+u) = 132, which simplifies to t + u = 12. Once the condition reduces to a single equation in two digits, Tool #13 (Work Systematically) takes over: list every (t,u) with t + u = 12, 1 ≤ t ≤ 9, 0 ≤ u ≤ 9, and count what survives.
Write the digits in place value
Let t be the tens digit and u the units digit, so N = 10t + u and its reversal is 10u + t.
Place value (Grade 4) says the tens digit contributes ten times its face value, and the ones digit contributes its face value.
4.NBT.A.1Look For A PatternTranslate the condition
Adding N and its reversal and setting the sum to 132 gives (10t + u) + (10u + t) = 132.
Writing a word condition as an algebraic equation is Grade 6 expressions-and-equations work.
6.EE.A.2Look For A PatternReduce to the digit sum
Combine like terms to 11(t + u) = 132, then divide by 11 to get t + u = 12.
Grouping 11t + 11u as 11(t+u) is the distributive property — a Grade 6 "equivalent expressions" move.
If a two-digit number added to the number with its digits reversed equals 132, then the tens digit and the units digit must add up to 12.
▸ Why?
Adding the number and its reversal and grouping the matching digits gives eleven of the tens digit plus eleven of the units digit, which is eleven times the sum of the two digits.
▸ Why?
By place value the number is 10 × (tens digit) + (units digit) and the reversal is 10 × (units digit) + (tens digit), because the tens place is worth ten times its face value.
▸ Why?
Ten of a digit plus one more of it is eleven of it, and eleven of the tens digit plus eleven of the units digit is eleven of their combined sum, since a shared factor can be pulled out of a sum.
▸ Why?
Since eleven times the digit sum equals 132, the digit sum has to be 132 ÷ 11 = 12, because dividing by eleven undoes multiplying by eleven.
Apply the digit limits
With t from 1 to 9 and u = 12 - t, keep only the pairs where u is also a valid digit (0 to 9).
Substituting each candidate t into t + u = 12 and checking the digit range is the Grade 6 idea of "which values make the equation true."
6.EE.B.5Convert To AlgebraCount the valid pairs
Each valid pair gives one number N = 10t + u, so counting them gives 7 numbers — answer (B).
Counting the items in a generated list is the Grade 4 "analyze a pattern" finishing step.
4.OA.C.5Convert To AlgebraOnce you write a two-digit number as 10t + u, the whole problem turns into a Grade 6 equation t + u = 12 — then you just list the digit pairs that work!
- Write the digits in place value
- Translate the condition
- Reduce to the digit sum
- Apply the digit limits
- Count the valid pairs
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