2025 AMC 10A Problems/Problem 19

(Problem goes here)

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.

~Squidget(note: this solution is incomplete)

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2025 AMC 10A Problems/Problem 19

Solution 1

Solution 2