AMC 10 · 2008 · #21
Grade 7 countingPick an answer.
Counting the heavy-tailed rows one by one means checking 120 rows, but the three classes less-than, equal, and greater-than are not independent: swapping the front pair with the back pair turns a less-than row into a greater-than row and back again. That pairing forces the two outer classes to have the same size, so the whole problem collapses to counting the small middle class where the two sums tie. Counting the ties is a short, finite listing job once the total 15 is used to pin down which middle entries are even possible.
Sort all 120 rows into three classes
Every ordering falls into one of three classes.
Two numbers can only relate in three ways, so the whole set of rows splits into three piles that account for everything.
7.SP.C.8Change Focus Count The ComplementSwap the ends to match L with G
Swapping the ends makes the first and last classes equal in size.
Trading the two ends of a row turns "front is lighter" into "front is heavier", so the two piles pair off exactly.
Trading the two ends of a row turns front-is-lighter into front-is-heavier, so the two piles pair off exactly.
▸ Why?
The swap sends each row of one pile to exactly one row of the other and back again, so the piles are the same size.
▸ Why?
Every row falls into exactly one of the three classes, so the three counts add up to the whole.
Use the total 15 to pin the middle
The total forces the middle entry to be odd.
An odd total minus an even chunk leaves an odd leftover, so the middle number cannot be even.
2.OA.C.3Organize Information In More WaysList the balanced splits
Listing the balanced splits gives just three.
Once the middle number is fixed, there are only three ways to pair up what is left, so checking them all is quick.
7.SP.C.8Make A Systematic ListTurn each split into row counts
Each split yields several orderings.
Shuffling numbers inside a pair never changes its sum, so every such shuffle keeps the tie and must be counted.
3.OA.A.1Make A Systematic ListSolve for L and read the choice
Subtracting the ties and halving gives 48, choice (C).
Two equal piles plus a counted middle pile fill up 120, so one subtraction and one halving finish it.
6.EE.B.5Change Focus Count The ComplementWhen two cases mirror each other, count only the cases stuck in the middle and let subtraction hand you the rest.
- Sort all 120 rows into three classes
- Swap the ends to match L with G
- Use the total 15 to pin the middle
- List the balanced splits
- Turn each split into row counts
- Solve for L and read the choice