Competition · AMC preparation · step 4 of 4
AMC 8 · 2006 · #24
Grade 6 number-theoryPick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The number CDCD is just the two-digit block CD written twice. Tool #5 (Look for a Pattern) spots that this repetition is the same trick as abab = ab × 101 (similar to how aa = a × 11). Once we name that pattern, Tool #13 (Convert to Algebra) turns the multiplication line into a one-line equation ABA × CD = CD × 101, which we can divide by CD to read ABA directly. No guessing needed.
Spot the repeating block
Spot that CDCD is the block CD copied twice, so place value expands it to CD × 101.
Repeating a 2-digit block in a 4-digit slot multiplies the block by 101 — the same Grade 5 place-value idea that makes aa = a · 11.
The four-digit number CDCD is exactly the two-digit block CD multiplied by 101.
▸ Why?
CDCD splits with no overlap into its top half CD in the thousands-and-hundreds places and its bottom half CD in the tens-and-ones places, and those two parts add back to the whole.
▸ Why?
A multi-digit number is just what each of its places is worth added together, so cutting it between the top two places and the bottom two places gives two pieces that sum back to the original number.
▸ Why?
The top copy of CD sits two places higher than the bottom copy, so it is worth 100 times that block while the bottom copy is worth the block once: CDCD = CD · 100 + CD.
▸ Why?
Each place is worth ten times the place to its right, so sliding the same digits two places left multiplies their value by ten twice, that is by 100.
▸ Why?
Both terms carry the same factor CD, so it pulls out: CD · 100 + CD · 1 = CD·(100 + 1) = CD · 101.
▸ Why?
A factor shared by every part of a sum can be taken outside the sum, since the same amount is being counted either way, only regrouped.
Write the multiplication as an equation
Turn the multiplication into an equation and substitute Step 1: ABA × CD = CD × 101.
Naming the unknown 3-digit number as ABA and rewriting the column-multiplication as a single equation is the Grade 6 "variables in expressions" move.
6.EE.A.2Convert To AlgebraCancel the common factor
Divide both sides by CD, which is nonzero since C ≠ 0, giving ABA = 101.
Dividing both sides by the same nonzero quantity is the Grade 6 one-step equation move — and it makes the answer fall out without solving for C or D at all.
6.EE.B.7Convert To AlgebraAdd A and B
Read the digits of 101 as A = 1 and B = 0, so A + B = 1.
Reading individual digits out of a 3-digit number is Grade 4 place-value, the same skill used to read 101 as one hundred and one.
4.NBT.A.2Look For A PatternWhen a block of digits repeats — like CDCD being CD written twice — it always factors out as that block times 101. Spot that pattern and the multiplication puzzle collapses to a Grade 6 one-step equation.
- Spot the repeating block
- Write the multiplication as an equation
- Cancel the common factor
- Add A and B
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