AMC 8 · 1999 · #1
Grade 6 arithmeticPick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The equation already tells us the final value is 5. Tool #11 (Work Backwards) lets us undo the known pieces — first simplify (2-1), then peel off the +4 - 1 — until only the mystery expression (6 ? 3) is left equal to a number. Once we know what (6 ? 3) must equal, Tool #6 (Guess and Check) tests each of the four operators on the small numbers 6 and 3 — a one-line check per option.
Simplify the second parenthesis first: (2 - 1) = 1, so the equation becomes (6 ? 3) + 4 - 1 = 5.
Order of operations says do the inside of parentheses first — a Grade 5 skill.
5.OA.A.1Work BackwardsCombine the known numbers: 4 - 1 = 3, so the equation now reads (6 ? 3) + 3 = 5.
4 - 1 = 3 is a Grade 3 within-100 subtraction fact.
3.NBT.A.2Work BackwardsSubtract 3 from both sides to isolate the mystery expression: (6 ? 3) = 2.
Subtracting the same value from both sides keeps the equation balanced — the Grade 6 inverse-operation move.
6.EE.B.7Work BackwardsTest each operator on 6 and 3 for the value 2: only 6 ÷ 3 = 2 works; the others give 18, 9, and 3.
All four checks use one-digit basic facts — Grade 3 fluency with multiplication and division within 100.
3.OA.C.7Guess And CheckClean up the parts you already know, see what the mystery piece has to equal, then try each operator. AMC 8 #1 is a Grade 6 balance-the-equation problem with a Grade 3 fact check at the end.