2025 AMC 10A Problems/Problem 19

Question

An array of numbers is constructed beginning with the numbers $-1$ , $3$ , and $1$ 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 $-1$ and $1$ , respectively.

$\large{-1}\qquad\large{3}\qquad\large{1}$ $\large{-1}\qquad\large{2}\qquad\large{4}\qquad\large{1}$ $\large{-1}\qquad\large{1}\qquad\large{6}\qquad\large{5}\qquad\large{1}$ If the process continues, one of the rows will sum to $12{,}288$ . In that row, what is the third number from the left?

$\textbf{(A)} \: -29\qquad\textbf{(B)} \: -21\qquad\textbf{(C)} \: -14\qquad\textbf{(D)} \: -18\qquad\textbf{(E)} \: -3$

Solution 1

Consider the polynomial $f(x) = -x^2+3x+1.$ When we multiply this polynomial by $x+1,$ we are essentially doing the operation given in the problem (When we multiply $p(x)$ by $x+1,$ a term of degree $d$ in the yielded expression is the sum of $1\cdot(\text{degree d})$ and $x\cdot(\text{degree d-1})$ in $p(x)$ This effect is visible in Pascal's Triangle). So, if we let the coefficients of $f(x)$ be the zero row of the array, then the $n^{th}$ row is just the coefficients of $f(x)(x+1)^n.$ The next thing to note is that the sum of the coefficients in any polynomial $p(x)$ is just $p(1).$ Therefore, the sum of the entries in the $n^{th}$ row of the array is $f(1)(1+1)^n=3*2^n.$ Letting this equal $12288,$ we get $n=12.$ We are looking for the $3^{rd}$ term in the $12^{th}$ row. The $12^{th}$ row is given by the coefficients of $f(x)(x+1)^{12}=(-x^2+3x+1)(x+1)^{12}.$ Since the degree of the resulting expression is $14,$ the third term in the row is just the coefficient of $x^{12}$ in the expression, which is $-\dbinom{12}{10}+3\dbinom{12}{11}+1=\boxed{\text{(A) }-29}.$

~Tacos_are_yummy_1

Solution 2

If we take a look at the first few rows, we notice that the sum of the terms in each row $n$ is equal to the twice the sum of row $n-1$ . We note the first row is $3$ so recognize $12,228$ must be equal to $3$ times a power of $2$ . $12,288=3\cdot 4096=3 \cdot 2^{12}$ . Therefore, we are looking for the $3$ rd term from the left in the $13$ th row. From here, we

~Squidget(note: this solution is incomplete, will complete soon)

Solution 3 - ⚡ Fast

Add all the numbers up on the first row. You get $3.$ Now, add all the numbers up on the second row. You get $6.$ Notice that as the rows keep going, the sum of the numbers in the row keep doubling. When you repeat this process, you realize that you reach $12288$ on the $13^{th}$ row.

Knowing this, we can use this pattern to quickly find the solution. We know that the first number will always be $-1$ , so we can ignore that. Knowing this, all you have to do now is to add up the second and third numbers $13$ times. This is already done for us three times, so we just have to do it ten more. Note that you do not have to do this process for all the numbers, only the second and third.

Doing this 13 times (since $\frac{12288}{3}=4096=2^{12}=2^{13-1}$ ), you get the following string of numbers (starting from the first one): $1, 4, 6, 7, 7, 6, 4, 1, -3, -8, -14, -21,$ and finally, $\boxed{\text{(A) }-29}.$

~i_am_not_suk_at_math (saharshdevaraju 21:19, 6 November 2025 (EST)saharshdevaraju)

~minor edits by i_am_not_suk_at_math (saharshdevaraju 21:19, 6 November 2025 (EST)saharshdevaraju)

~minor edits by iiiiiizh

~minor LaTeX edits by zoyashaikh

~minor LaTeX edits by yogert2

Solution 4

From adding the sum of all terms in the first few rows, we see that $S(r(1))=3, S(r(2))=6, S(r(3))=12, S(r(4))=24,...$ and so on. We can deduce that $S(r(n))=3 \cdot 2^{n-1}$ is the sum of all numbers in row n. Now, set $3 \cdot 2^{n-1} = 12288$ so we have $2^{n-1} = 4096 = 2^{12}$ . It follows that $n=13$ . Now let $a_n$ be the 2nd number of each row, and $b_n$ be the 3rd number of each row. Since, the first number of each row is -1, $a_n=a_{n-1}-1$ . Additionally, $b_n = b_{n-1}+a_{n-1}$ .

\begin{array}{|c|c|} \hline a_n & b_n \\ \hline a_1 = 3 & b_1 = 1 \\ a_2 = 2 & b_2 = 4 \\ a_3 = 1 & b_3 = 6 \\ a_4 = 0 & b_4 = 7 \\ a_5 = -1 & b_5 = 7 \\ a_6 = -2 & b_6 = 6 \\ a_7 = -3 & b_7 = 4 \\ a_8 = -4 & b_8 = 1 \\ a_9 = -5 & b_9 = -3 \\ a_{10} = -6 & b_{10} = -8 \\ a_{11} = -7 & b_{11} = -14 \\ a_{12} = -8 & b_{12} = -21 \\ a_{13} = -9 & b_{13} = -29 \\ \hline \end{array}

We are asked to compute $b_{13} = \boxed{\text{(A) }-29}.$

~hxve

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 21:29, 6 November 2025 (EST)saharshdevaraju)

Solution 5 (Wish I used this)

So the sum is doubling every time, starting with 3.

$12288 = 3 * 2^{12}$

So in the 13th row we go, so we eventually get that the 3rd number is $\boxed{\text{(A)}-29}$

Go check out solution 1, looks interesting.

~Aarav22

Video Solution (In 2 Mins)

https://youtu.be/yD1EcmcjZGU?si=-UoUuK-GQolFhu9t ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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