Competition · AMC preparation · step 4 of 4
AMC 8 · 1999 · #11
Grade 6 arithmetic
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Drawing the plus sign (Tool #1) makes the key feature visible: the center square sits inside the row AND inside the column. If you add the row sum and the column sum, every outside number is counted once but the center number is counted twice. That gives the invariant equation 2S = 35 + center — Tool #11 (Find an Invariant). Since 35 is fixed, the equation pins S entirely to the center value, so maximizing S just means making the center as large as possible. No guessing of arm placements is needed.
Mark the shared square
Sketch the plus sign: the row (left, center, right) and the column (top, center, bottom) share the center square.
Drawing the cross shape lets you see at a glance that one square does double duty. Without the picture, it's easy to forget the center counts twice.
4.OA.A.3Draw A DiagramAdd the row and column sums
Add row + column: the four arm numbers count once, the center counts twice, and all five total 35.
Counting each square's contribution to (row sum) + (column sum) shows the center is the only square hit twice. The fixed total 35 is the invariant — it doesn't depend on which number goes where.
Adding the row's three numbers to the column's three numbers equals the total of all five numbers plus one extra copy of the center number: S + S = 35 + center.
▸ Why?
The row total is its three squares added up and the column total is its three squares added up, so setting the two totals together lists all six filled slots — the four arm numbers once each, and the center number twice, since the center square is a part of the row and also a part of the column.
▸ Why?
That list of six numbers can be sorted into the four arm numbers plus one center number — which together are exactly the five given numbers — with a single center number left over, so the combined total is 35 + center.
▸ Why?
The four arm numbers together with one center number are just the five numbers 1, 4, 7, 10, 13 placed in the squares, and those five add to a fixed 35 wherever each one sits.
▸ Why?
You may reorder and regroup the six terms freely while adding, so pulling the four arm numbers and one center number together into 35 and leaving the other center number aside does not change the total.
▸ Why?
Changing the order in which you add the numbers never changes the total.
▸ Why?
Changing which numbers you group and add first never changes the total.
Solve for the common sum
Solve the invariant equation for S. Divide both sides by 2.
Now S depends only on the center number. The four arm squares can be arranged in many ways, but the common sum is locked once the center is chosen.
6.EE.B.7Work BackwardsMake the center as large as possible
Bigger center means bigger S, so put the largest number 13 in the center: S = = 24 (48 is even, so S is whole).
Bigger center means bigger S, so go for the biggest number. Need to be sure 35 + center is even — it is for 13, and that gives a clean integer answer.
6.EE.B.7Work BackwardsPlace the rest to check
Split the rest {1, 4, 7, 10} into pairs summing 11: (1, 10) and (4, 7). Then row 1+13+10 and column 4+13+7 both equal 24 → (D).
The arrangement exists, so S = 24 is actually reachable — not just an upper bound from the equation.
4.OA.A.3Draw A DiagramThe center square is counted twice when you add the row and the column, so 2S = 35 + center. To make S biggest, put the biggest number (13) in the center — that gives S = 24.
- Mark the shared square
- Add the row and column sums
- Solve for the common sum
- Make the center as large as possible
- Place the rest to check
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