New Light Math

I was working through some Chinese Remainder Theorem (CRT) problems today (9/24/22), using a new technique introduced in the UKMT Introduction to Number Theory Book. And I noticed a pattern: \[9a \equiv 1 \pmod 7 \rightarrow a = 4\] \[7b \equiv 1 \pmod 9 \rightarrow b = 4\] Which is cool. But it continues: \[7c \equiv 1 \pmod 5 \rightarrow c = 3\] \[5d \equiv 1 \pmod 7 \rightarrow d = 3\] And: \[35e \equiv 1 \pmod{33} \rightarrow e = 17\] \[33f \equiv 1 \pmod {35} \rightarrow f = 17\] Each modular inverse for these pairs were the same! I formed a conjecture, and proved it as a new theorem: Theorem : The solutions $a, b$ to the modular equivalences \[(n-1)a \equiv 1 \pmod {n+1}\] \[(n+1)b \equiv 1 \pmod {n-1}\] for positive even integers $n$ satisfy $a = b = \frac{n}{2}$. Update  (12/25/23): More intuitive understanding of Chinese Remainder Theorem (CRT) process. I was looking back at this post, and realized that I had forgotten how the Chinese Remainder Theorem process wo...

Fall 2021 AMC 12B #24 Problem:  Triangle $ABC$ has side lengths $AB = 11, BC=24$, and $CA = 20$. The bisector of $\angle{BAC}$ intersects $\overline{BC}$ in point $D$, and intersects the circumcircle of $\triangle{ABC}$ in point $E \ne A$. The circumcircle of $\triangle{BED}$ intersects the line $AB$ in points $B$ and $F \ne B$. What is $CF$? Solution:  We can create a diagram to try to find unique relationships in this problem. By angle chasing in the cyclic quadrilateral $ABEC$ given that $AE$ is the angle bisector of $\angle{BAC}$ we can find that $\angle{BAE} = \angle{CAE} = \angle{BCE} = \angle{CBE}$. This means that triangle $BCE$ is isosceles and $BE = CE$, which may be useful later on. Now, we can focus on $CF$ , which is the length we want to find. We can see that $ACF$ has a cevian $CB$, so we can use Stewart's Theorem (we will "derive" it with the law of cosines here) to solve for CF in terms of other side lengths. We can let $\angle{ABC} = \theta$ which means...

Fall 2021 AMC 12B #15 Problem:  Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$-sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c$? Solution:  We can see that the resulting shape is a "regular" 24-sided polygon with equal side lengths, and equal interior angles. So we can take a "petal" of the polygon that consists of 3 adjacent vertices, find the area, and multiply by 12. The key here is to find the area of the petal by subtracting areas from a corner of the original square paper. From this diagram of a petal inside a $3 \times 3$ corner of a square, we can see that each petal has an area...

Fall 2021 AMC 12B #5 Problem:  Call a fraction $\frac{a}{b}$, not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$. How many distinct integers can be written as the sum of two, not necessarily different, special fractions? Solution:  We can list out all the special fractions: $\{\frac{1}{14},\frac{2}{13}, \frac{3}{12}, \frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3}, \frac{13}{2}, \frac{14}{1}\}$. Simplifying, we have the fractions $\{\frac{1}{14},\frac{2}{13}, \frac{1}{4}, \frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},\frac{8}{7},\frac{3}{2},2,\frac{11}{4},4, \frac{13}{2}, 14\}$. We see that we can only add fractions with the same denominator , so we only need to focus on $\{\frac{1}{4},\frac{1}{2},\frac{2}{3},\frac{3}{2},2,\frac{11}{4},4, \frac{13}{2}, 14\}$. From the whole numbers $\{2,4,14\}$, we can achieve $4, 6, 16, 8, 18, 28$. From the fractions with deno...

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...

Choosing subsets without consecutive integers (in one form or another) is a problem that comes up quite frequently in harder problems. This approach to solving those problems is both simple and useful. To find a possible subset of {1, 2, 3, ..., x} that does not contain consecutive integers, we can first note that integers of a subset must have a "space" or integer between them. Next, the key to this approach is to think of all x integers in the set as dots . By ordering specific dots in a sequence, we can map the chosen (or red) dots to the chosen integers in the subset. Now we have that chosen dots are red, so let the "non-chosen" dots be green. But all the orderings include consecutive integers, so we need to remove some "buffer" dots that can later be added between the chosen dots. So suppose we need to find the total number of 3 integer subsets of {1, 2, 3, ..., 7} such that no two integers are consecutive. Then we have 2 "buffer dots" to be...

2006 AMC 12A #8 Problem: How many sets of two or more consecutive positive integers have a sum of 15? Solution: It is simple to see there are 3 total ways: {1, 2, 3, 4, 5} | {4, 5, 6} | {7, 8}. But why? Is there a way to generalize this? Yes! The number of sets of two or more positive integers that have a sum of k is equal to the number of positive odd divisors of k excluding 1. Let's break this down with an example. Take 369 = 3^2 * 41. From above, we have 5 ways. To find such sets, we take the factors of 369 and let them be our median of the set. Then, (regardless if any of the integers in the set are negative) we have a feasible set. First, 123 will be our median. Our set is: {122, 123, 124}. This set has 3 numbers and needs no manipulation. Next, our median equals 41. Our set becomes longer: {37, 38, 39, 40, 41, 42, 43, 44, 45}. This also doesn't require manipulation. Now, we have 9 as our median. Our set is then: {-11, -10 , -9, ... 9, 10, 11, ... 29}. However, this set ...

2019 AMC 10B #1 Problem: Alicia had two containers. The first was $\frac{5}{6}$  full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\frac{3}{4}$  full of water. What is the ratio of the volume of the first container to the volume of the second container? Solution: Let the volume of the first container be equal to X. Similarly, define Y to be the volume of the second container. From the problem, we see that  $\frac{5}{6}$ X = $\frac{3}{4}$Y. Solving, we get X/Y = 9/10 . 2019 AMC 10B #12 Problem: What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than 2019? Solution:  We see that the largest digit in any base-seven representation of a positive number is 6, so we can maximize the number of 6's in the base-seven representation. We know that $666_7 = 6*(7^0+7^1+7^2) = 342_10$. So, we have 2019-342 = 1677 fo...

2020 AMC 10A #22 Problem: For how many positive integers $n \leq 1000$  is \[ \lfloor{\frac{998}{n}}\rfloor + \lfloor{\frac{999}{n}}\rfloor + \lfloor{\frac{1000}{n}}\rfloor \] not divisible by 3 ? (Recall that $ \lfloor x \rfloor $  is the greatest integer less than or equal to $x$ .) Solution: Observe that the 3 values $\lfloor{\frac{998}{n}}\rfloor$,  $\lfloor \frac{999}{n} \rfloor $ , and  $\lfloor{\frac{1000}{n}}\rfloor$ must have exactly 1 of them that is not equal to the others, to satisfy the condition that the expression  \[ \lfloor{\frac{998}{n}}\rfloor + \lfloor{\frac{999}{n}}\rfloor + \lfloor{\frac{1000}{n}}\rfloor \] is not divisible by 3. Note that the "turning points", or values of n that produce different values for $\lfloor{\frac{998}{n}}\rfloor$,  $\lfloor \frac{999}{n} \rfloor $ , and  $\lfloor{\frac{1000}{n}}\rfloor$ are the factors of 999 and 1000 . We realize that 998 has 4 factors: 1, 2, 499, 998. Note that when $n=1$, 499...

Math Prize for Girls 2019 #10 Problem:  A 1*5 rectangle is split into five unit squares (cells) numbered 1 through 5 from left to right. A frog starts at cell 1. Every second it jumps from its current cell to one of the adjacent cells. The frog makes exactly 15 jumps. How many paths can the frog take to finish at cell 5? Solution:  When we see such a problem, we automatically think of using some counting of permutations to help solve this, but actually, all we need is addition.        1        2        3        4       5 Step 1        1 Step 2       1       1 Step 3       2       1 Step 4       2       3       1 Step 5       5       4 Step 6       5       9      ...

Q: If x is between 0 and 1, what is the probability that the greatest  integer ≤ log n (1/x) is odd, in terms of n. Definition of log n (1/x) = y is n y = 1/x. S: For the greatest integer ≤ log n (1/x) to equal 1, we must have a value  between 1/n and 1/n 2 . Then for the greatest integer ≤ log n (1/x) to equal 3, we must have a value between 1/n 3 and 1/n 4 . This process continues and we  are left with a geometric sequence:    (n-1)/n 2 + (n-1)/n 4 + (n-1)/n 6 + (n-1)/n 8 ….. = (n-1)/n 2    (n 2 -1)/n 2 = __ (n-1)__    (n+1)(n-1) =  _ 1 _   (n+1) Q: What is the remainder when (1)(1+2)(1+2+3)(1+2+3+4)...(1+2+3..+58+59) is divided by 61? S: We can rewrite this as:  1*2*2*3*3*4*4*5*5*6*...*58*59*59*60 2^59 We can then change this to: 59!*60!   2^59 By Wilson's formula, for any prime p , (p-1)! + 1 ≡ 0 (mod p) We can then ...