AMC 10 · 2003 · #3
Grade 7 algebraPick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The five unknown evens are all tied to one number — the smallest — so naming it x (Tool #4) turns five mystery values into one expression 5x+20. Tool #5 (Look for a Pattern) collapses the other side fast: the first n odd numbers always total n², so eight of them give 64 without adding one by one. Tool #13 (Convert to Algebra) then turns the sentence '4 less than' into a single equation to solve.
Add the first 8 odd numbers
The first 8 odd numbers 1,3,5,…,15 total 64 — the first n odds always make n².
Odd numbers stacked from 1 build a perfect square, so n of them make an n × n block.
Odd numbers stacked from one build a perfect square, so a run of them makes a square block.
▸ Why?
The odd numbers climb by the same fixed step, so the list is evenly spaced.
▸ Why?
Pairing the first with the last gives the same total as pairing inward, so the sum is the count times the middle.
Name the smallest even integer
Call the smallest even integer x; the five are x, x+2, x+4, x+6, x+8, adding to 5x+20.
One name for the smallest value forces every other value, so five unknowns shrink to one.
6.EE.B.6Introduce A VariableTurn the sentence into an equation
'4 less than' the odd total of 64 gives 64-4, so the equation is 5x+20 = 60.
'4 less than 64' is just 64-4, and the word 'is' becomes an equals sign.
7.EE.B.4Convert To AlgebraSolve and check
Subtract 20, then divide by 5: x = 8 — choice (B). Check: 8+10+12+14+16 = 60.
Undo the equation in reverse — remove the +20, then the ×5 — to uncover x.
7.EE.B.4Introduce A VariableName the smallest number x, write the rest as steps of 2 above it, and one clean equation does the rest.
- Add the first 8 odd numbers
- Name the smallest even integer
- Turn the sentence into an equation
- Solve and check