New Light Math

Fall 2021 AMC 12A #25 Problem:  Let $m\geq 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\leq a_i \leq m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$. There is a polynomial $q(x) = c_3x^3+c_2x^2+c_1x+c_0$ such that $D(m) = q(m)$ for all odd integers $m\geq 5$. What is $c_1$? Solution:  First, we must find a way to account for the number of ordered quadruples $(a_1, a_2, a_3, a_4)$ such that $m$ divides $a_1+a_2+a_3+a_4$. We know that there are $(m)(m-1)(m-2)(m-3)$ total ordered quadruples since all elements must be distinct. First we can list out the sums of $m$ quadruples of the form $(a_1+n, a_2+n, a_3+n, a_4+n)$ where $n$ takes on the values from $0$ to $m-1$ and all elements of the quadruple are modulo $m$. This means that the $m$ quadruples have the same difference between consecutive elements modulo $m$. If we let the sum $S = a_1+a_2+a_3+a_4$, we have $(a_1, a_2, a_3, a_4) \Rightarr...

Not following through with a problem may be the most difficult pitfall to avoid. Try the problem below (it's casework) and try to follow through with all cases. Fall 2021 AMC 12A #24 Problem:  Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}$. In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a$. What is the sum of all possible values of $a$? Solution:  We see that the sides of $ABCD$ must have side lengths $x, x+d, x+2d, x+3d$ to form an arithmetic sequence with difference $d \geq 0$. Since $\overline{AB}$ has the maximum length, we have that $AB = x+3d = 18$, with $3! = 6$ ways to arrange the other sides of length $x, x+d$, and $x+2d$. From the diagram, we can relate the length of $AB$ to the length of $CD$ because the distance between the two parallel sides is the same at all points. So, our equation becom...

Here's an interesting quadratic polynomial problem that utilizes more than one quadratic! Try it on your own first before following the solution. Fall 2021 AMC 12A #23 Problem:  A quadratic polynomial with real coefficients and leading coefficient $1$ is called disrespectful if the equation $p(p(x))=0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$? Solution:  First, we can let the two roots of $p(x)$ be $r$ and $s$. We see that in order for $p(p(x))$ to equal zero, $p(x)$ must be equal to either $r$ or $s$ . So, if we let $p(x) = (x-r)(x-s)$, we have that the solutions to $p(p(x))=0$ correspond to the solutions to $x^2-(r+s)x+rs = r$ and $x^2-(r+s)x+rs = s$. If we let the three numbers that satisfy $p(p(x))=0$ be $m, n, q$, we can see that either one of the quadratic equations above has roots $m$ and $n$, and the...

Notable Early Problems from Fall 2021 AMC 12A Fall  2021  AMC 12A #8 Problem : Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40$. What is the value of $\frac{N}{M}$? Solution:  We can see that $N$ contains all the factors of the $M$, which in turn contains all the factors of the numbers from 10 to 30. So, we must consider the additional terms $32,33,34,35,36,37,38,39,$ and $40$ to determine what additional factors $N$ has other than $M$ . Beginning with $32$, we see that $32 = 2^5$, but $M$ only contains $16 = 2^4$, so $N$ has one extra factor of $2$ than $M$. $33$ is already included as $3*11$ is a factor of two relatively prime numbers, $18*11$. $34=2*17$, which is included in $10*17$. $35=5*7$ is included in $10*21$ and $36=4*9$ is included in $16*27$. $37$ is prime, so $N$ has an additional factor of $37$ as well. Finally, $38=2*19$ is included in $1...

2016 AMC 10A #17 Problem: Let N be a positive multiple of 5. One red ball and N green balls are arranged in a line in random order. Let P(N) be the probability that at least 3/5 of the green balls are on the same side of the red ball. Observe that P(5) = 1 and that P(N) approaches 4/5 as N grows large. What is the sum of the digits of the least value of N such that P(N) < 321/400? Generalized Problem : Let N be a multiple of odd positive integer M. One red ball and N green balls are arranged in a line in random order. Let P(N) be the probability that at least $\frac{\frac{M+1}{2}}{M}$ of the green balls are on the same side of the red ball. What is P(N) in terms of M? Solution:  Let L be $\frac{N}{M}$, or $N=L*M$. First, we must note that we can first place all the N green balls in a line, with N+1 places to place the red ball, with equal probability . Then, all we need to find is the number of invalid positions and subtract it from the total N+1 positions of the red b...