Competition · AMC preparation · step 4 of 4
AMC 8 · 2003 · #14
Grade 4 number-theoryPick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A cryptarithm is one big puzzle, but column addition hands us four ready-made subproblems (Tool #7): the thousands column, the hundreds column, the units column, and the tens column. Solve them in the order that locks in the most letters first — thousands → hundreds → units → tens. Once T, F, O, R are pinned down, the tens column reduces to choosing W from a tiny list. There Tool #3 (Eliminate Possibilities) takes over: test W = 0, 1, 2, 3, 4 against the "all digits different" rule and cross out the bad ones.
Solve the thousands column
Thousands column: the extra leading digit is a carry, and doubling a three-digit number caps that carry, so F = 1.
Grade 4 place-value reasoning: the digit that "falls out the top" of a column is the carry, and doubling a 3-digit number can carry at most 1.
4.NBT.A.2Identify SubproblemsSolve the hundreds column
Hundreds column: 14 + c₂ = 10 + O gives O = 4 + c₂, and the even-O condition selects O = 4 with carry c₂ = 0.
Grade 4 standard-algorithm addition: a column's digit equals the column sum minus 10 · (carry out). The even/odd filter picks one case.
In the hundreds column, 7 + 7 plus the carry from the tens must read as a leading ten followed by O, and the rule that O is even pins O = 4.
▸ Why?
Reaching ten in the hundreds column bundles up into a brand-new thousands place — the single leading digit — while the remainder stays as O, so the column reads 14 + c = 10 + O, giving O = 4 + c.
▸ Why?
A place holds only up to nine, so the tenth unit bundles up into the next place; that is why the overflow shows up as one new leading digit and the leftover part stays behind as O.
▸ Why?
The carry c arriving from the tens column is 0 or 1, because doubling two single digits (plus at most a 1 carried in) can bundle at most one new ten upward.
▸ Why?
Since O = 4 + c with c equal to 0 or 1, the only options are O = 4 or O = 5, and the given that O is even discards the odd 5.
▸ Why?
4 and 5 are consecutive whole numbers, so they carry opposite parity — one splits into two equal whole groups and the very next number cannot — meaning exactly one of 4 and 5 is even.
Solve the units column
Units column: O = 4 gives O + O = 8, still under ten, so R = 8 and nothing carries into the tens (c₁ = 0).
No regrouping needed here — the column sum is a single digit, so R is exactly that digit and nothing carries left.
4.NBT.B.4Identify SubproblemsSolve the tens column
Tens column: with both carries 0, W + W = U and 2W under ten; only W = 3 gives a U (6) that no other letter already uses.
Grade 4 "reason about whole numbers": W must double to a single digit and must keep every letter on a different value. Five candidates, one survivor.
4.OA.B.4Eliminate PossibilitiesA cryptarithm is just column addition in disguise. Solve one column at a time, lock in F, then O, then R, and the tens column hands you a short list for W — only W = 3 keeps every letter on a different digit.
- Solve the thousands column
- Solve the hundreds column
- Solve the units column
- Solve the tens column
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