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2025 AMC 10A Problem 19

P16-20~5 minPrint
Sequences & SeriesAlgebraic ManipulationCompute small cases, spot the pattern, generalizeDefine states/recurrence and iterate

An array of numbers is constructed beginning with the numbers −131-1\qquad3\qquad1 in the top row. Each adjacent pair of numbers is summed to produce a number in the next row. Each row begins and ends with −1-1 and 11 , respectively.

−131−1241−11651\begin{array}{ccccccccc} &&-1&&3&&1&&\\ &-1&&2&&4&&1&\\ -1&&1&&6&&5&&1\\ \end{array}

If the process continues, one of the rows will sum to 12,28812,288 . In that row, what is the third number from the left?

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Key insight

Each row adds up to twice the row above, so 12,288 is the sum of row 13. Then follow only the second and third numbers down to that row.

Solution

The question has two parts: first find which row adds up to 12,28812{,}288, then find the third number of that row.

Part 1: which row.

1. Start by adding up the three rows in the picture:
- Row 1: −1+3+1=3-1 + 3 + 1 = 3
- Row 2: −1+2+4+1=6-1 + 2 + 4 + 1 = 6
- Row 3: −1+1+6+5+1=12-1 + 1 + 6 + 5 + 1 = 12

The sum doubles each time. Before trusting that, check why it happens.

2. Look at how row 3 is built from row 2, which is −1,2,4,1-1, 2, 4, 1. The middle numbers of row 3 are the sums of neighbours: −1+2-1+2, 2+42+4, 4+14+1. The inside numbers 22 and 44 each get used twice, and the end numbers −1-1 and 11 get used once each. So these three sums add up to
2×6−(−1)−1=12.2 \times 6 - (-1) - 1 = 12.
3. Then row 3 gets a new −1-1 at the start and a new 11 at the end. Those two cancel, so the row sum is exactly 2×6=122 \times 6 = 12.
4. The same thing happens in every row, because every row starts with −1-1 and ends with 11. So the sums really do double: 3,6,12,24,…3, 6, 12, 24, \ldots Row nn adds up to 3×2 n−13 \times 2^{\,n-1} (row 1 is 3×13 \times 1, row 2 is 3×23 \times 2, row 3 is 3×43 \times 4).
5. Now 12,288÷3=4096=21212{,}288 \div 3 = 4096 = 2^{12}. So n−1=12n - 1 = 12, which means the row is row 13, not row 12.

Part 2: the third number of row 13.

6. Writing out whole rows down to row 13 would be slow, since row 13 has 1515 numbers. Luckily the first three numbers of a row depend only on the first three numbers of the row above:
- the first number is always −1-1;
- the new second number is −1+(old second number)-1 + (\text{old second number});
- the new third number is (old second number)+(old third number)(\text{old second number}) + (\text{old third number}).
7. So the second number goes down by 11 in every row, and the third number grows by adding the old second number. Make a table, starting from row 1 (33 and 11):

Row2nd number3rd number
13311
2221+3=41 + 3 = 4
3114+2=64 + 2 = 6
4006+1=76 + 1 = 7
5−1-17+0=77 + 0 = 7
6−2-27−1=67 - 1 = 6
7−3-36−2=46 - 2 = 4
8−4-44−3=14 - 3 = 1
9−5-51−4=−31 - 4 = -3
10−6-6−3−5=−8-3 - 5 = -8
11−7-7−8−6=−14-8 - 6 = -14
12−8-8−14−7=−21-14 - 7 = -21
13−9-9−21−8=−29-21 - 8 = -29
  1. Rows 2 and 3 match the picture (2,42, 4 and 1,61, 6), so the table is set up correctly.
  2. The third number of row 13 is −29-29.

The answer is (A) −29\boxed{\textbf{(A)}\ -29}.

Why this works

When each row is built from the row above, you rarely need whole rows. Pick out just the things the question asks about (here the row sum and the first three numbers) and find the rule for how each one changes from one row to the next. The sum doubles because every number is used twice except the ends, and the ends are set up to cancel. The second number drops by 11 each time, so the third number is just a running total.

Alternative approach

Skip the table and add in one go. The third number starts at 11 in row 1, and each step down adds the second number of the row above. To reach row 13 there are 1212 steps, which add the second numbers of rows 1 to 12:
3+2+1+0+(−1)+(−2)+⋯+(−8).3 + 2 + 1 + 0 + (-1) + (-2) + \cdots + (-8).
That is 1212 numbers going evenly from 33 down to −8-8. Their average is 3+(−8)2=−2.5\frac{3 + (-8)}{2} = -2.5, so their total is 12×(−2.5)=−3012 \times (-2.5) = -30. The third number of row 13 is 1+(−30)=−291 + (-30) = -29.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Which row. Let row nn be a1,a2,…,ama_1, a_2, \ldots, a_m with a1=−1a_1 = -1 and am=1a_m = 1, and let SnS_n be its sum. Row n+1n+1 consists of −1-1, then the m−1m-1 pairwise sums a1+a2,…,am−1+ama_1 + a_2, \ldots, a_{m-1} + a_m, then 11. In those pairwise sums every entry of row nn appears twice except the two ends, which appear once, so they total
2Sn−a1−am=2Sn+1−1=2Sn,2S_n - a_1 - a_m = 2S_n + 1 - 1 = 2S_n,
and the appended −1-1 and 11 cancel. Hence
Sn+1=2Sn.S_{n+1} = 2S_n.
With S1=−1+3+1=3S_1 = -1 + 3 + 1 = 3 we get Sn=3⋅2 n−1S_n = 3 \cdot 2^{\,n-1}; the printed rows check this (3,6,123, 6, 12). Since 12,288=3⋅4096=3⋅21212{,}288 = 3 \cdot 4096 = 3 \cdot 2^{12}, we need n−1=12n - 1 = 12, so it is row 1313.

The third entry. Write bnb_n and cnc_n for the second and third entries of row nn. From the construction,
bn+1=a1+bn=bn−1,cn+1=bn+cn.b_{n+1} = a_1 + b_n = b_n - 1, \qquad c_{n+1} = b_n + c_n.
Starting from b1=3b_1 = 3, the first recursion gives bn=4−nb_n = 4 - n. Then
cn=c1+∑j=1n−1bj=1+∑j=1n−1(4−j)=1+4(n−1)−(n−1)n2.c_n = c_1 + \sum_{j=1}^{n-1} b_j = 1 + \sum_{j=1}^{n-1}(4 - j) = 1 + 4(n-1) - \frac{(n-1)n}{2}.
Checking against the picture: c2=1+4−1=4c_2 = 1 + 4 - 1 = 4 and c3=1+8−3=6c_3 = 1 + 8 - 3 = 6, as printed.

For n=13n = 13,
c13=1+4⋅12−12⋅132=1+48−78=−29.c_{13} = 1 + 4 \cdot 12 - \frac{12 \cdot 13}{2} = 1 + 48 - 78 = -29.

The answer is (A) −29\boxed{\textbf{(A)}\ -29}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.