AMC 10 · 2008 · #2
Easy mode Grade 2Look at the 4 by 4 grid of numbers below. First, flip the second row so its numbers read right to left. Do the same to the fourth row. Now add the four numbers along each of the two diagonals. How much bigger is the larger diagonal sum than the smaller one?
18152229162331017244111825
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A $4\times4$ block of calendar dates is given. Reverse the left-to-right order of the numbers in the second row and in the fourth row, leaving the first and third rows alone. Then add the four numbers along each of the two diagonals and find the positive difference between those two diagonal sums.
Givens: The block reads row by row: $1,2,3,4$ then $8,9,10,11$ then $15,16,17,18$ then $22,23,24,25$; Only the second row and the fourth row get their numbers reversed left-to-right; Each diagonal has four numbers; the two diagonals are the top-left$\to$bottom-right and the top-right$\to$bottom-left; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $8$, (E) $10$
Unknowns: The positive difference between the two diagonal sums after the reversal
Understand
Restated: A $4\times4$ block of calendar dates is given. Reverse the left-to-right order of the numbers in the second row and in the fourth row, leaving the first and third rows alone. Then add the four numbers along each of the two diagonals and find the positive difference between those two diagonal sums.
Givens: The block reads row by row: $1,2,3,4$ then $8,9,10,11$ then $15,16,17,18$ then $22,23,24,25$; Only the second row and the fourth row get their numbers reversed left-to-right; Each diagonal has four numbers; the two diagonals are the top-left$\to$bottom-right and the top-right$\to$bottom-left; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $8$, (E) $10$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The whole difficulty is keeping track of which number sits in which cell after two rows flip, so Tool #1 (Draw a Diagram) — literally rewriting the block with rows 2 and 4 reversed — removes every chance of grabbing the wrong number off a diagonal. Once the new block is drawn, Tool #7 (Identify Subproblems) splits the job into three tiny, independent calculations: sum one diagonal, sum the other, subtract.
Execute — Answer: B
2.NBT.B.6 Step 1 Redraw the block after reversing
- Keep rows 1 and 3 the same.
- Reverse row 2 from $8,9,10,11$ to $11,10,9,8$, and reverse row 4 from $22,23,24,25$ to $25,24,23,22$.
- The new block, top to bottom, is: row 1 $= 1,2,3,4$; row 2 $= 11,10,9,8$; row 3 $= 15,16,17,18$; row 4 $= 25,24,23,22$.
💡 Draw the board exactly as it looks after the flip so you read real numbers off it, not remembered ones.
2.NBT.B.6 Step 2 Add the main diagonal
- The top-left$\to$bottom-right diagonal picks the first number of row 1, the second of row 2, the third of row 3, and the fourth of row 4: that is $1$, $10$, $17$, and $22$.
- Add them: $1+10+17+22=50$.
💡 Step down the diagonal one row at a time, moving one column right each row.
2.NBT.B.6 Step 3 Add the other diagonal
- The top-right$\to$bottom-left diagonal picks the fourth number of row 1, the third of row 2, the second of row 3, and the first of row 4: that is $4$, $9$, $16$, and $25$.
- Add them: $4+9+16+25=54$.
💡 This diagonal starts at the top-right corner and steps down-left, one column left each row.
2.NBT.B.5 Step 4 Take the positive difference
- The two diagonal sums are $50$ and $54$.
- The positive difference is the larger minus the smaller: $54-50=4$.
- So the answer is (B).
💡 Positive difference just means subtract the smaller sum from the larger one.
2.NBT.B.6 Keep rows 1 and 3 the same. Reverse row 2 from $8,9,10,11$ to $11,10,9,8$, and r 2.NBT.B.6 The top-left$\to$bottom-right diagonal picks the first number of row 1, the seco 2.NBT.B.6 The top-right$\to$bottom-left diagonal picks the fourth number of row 1, the thi 2.NBT.B.5 The two diagonal sums are $50$ and $54$. The positive difference is the larger m Review
Reasonableness: Before any reversal, both diagonals of this calendar block already sum to the same value $52$ (main $1+9+17+25=52$, anti $4+10+16+22=52$), so a difference can only come from the cells the flip actually moves. On the main diagonal the moved cells change $9\to10$ ($+1$) and $25\to22$ ($-3$), a net $-2$, giving $50$. On the other diagonal they change $10\to9$ ($-1$) and $22\to25$ ($+3$), a net $+2$, giving $54$. A shift of $-2$ against $+2$ makes a gap of exactly $4$, matching (B), and the answer is a small even number as all the choices are.
Alternative: Skip both full sums. Since the diagonals start equal at $52$, only the two swapped diagonal entries in each row matter. In row 2 the diagonal cell goes from $9$ up to $10$ on one diagonal but from $10$ down to $9$ on the other; in row 4 it goes from $25$ down to $22$ on one but from $22$ up to $25$ on the other. Each diagonal moves by $|1|+|3|$ split as $-2$ and $+2$, so they end $2+2=4$ apart.
CCSS standards used (min grade 2)
2.NBT.B.6Add up to four two-digit numbers using strategies based on place value and properties of operations (Adding the four numbers on each diagonal, $1+10+17+22=50$ and $4+9+16+25=54$.)2.NBT.B.5Fluently add and subtract within 100 using strategies based on place value and properties of operations (Computing the positive difference of the two diagonal sums, $54-50=4$.)
⭐ When numbers move around a grid, redraw the grid first and then read the diagonals off the new picture instead of your memory.
⭐ When numbers move around a grid, redraw the grid first and then read the diagonals off the new picture instead of your memory.
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