Number Theory: Proofs of AMC 10 Problems

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 for the leading digit, which is at most $4*7^3 = 1372$. Thus, we have the number $4666_7$, which yields a sum of $4+6+6+6= 22$.

We should also check whether we can achieve a higher sum if we increased the leading digit and decreased the other digits. The best we can get is $5566_7 = 2008_10$, which gives us the same sum of $22$, which is our answer.

2019 AMC 10B #18

Problem: Henry decides one morning to do a workout, and he walks $\frac{3}{4}$ of the way from his home to his gym. The gym is 2 kilometers away from Henry's home. At that point, he changes his mind and walks $\frac{3}{4}$ of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks $\frac{3}{4}$ of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked $\frac{3}{4}$ of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point $A$ kilometers from home and a point $B$ kilometers from home. What is $|A-B|$?

Solution: The key for this problem is to assume Henry is already at the "equilibrium position," where he walks back and forth between $A$ and $B$. Then, letting the distance $|A-B| = x$, we have that $HA = \frac{x}{3}$ where H is Henry's home since $\frac{AB}{HB} = \frac{3}{4}$. Similarly, since $\frac{AB}{AG} = \frac{3}{4}$, we get $BG = \frac{x}{3}$. So, the total distance $HG = \frac{5}{3}x = 2$, which means $x = |A-B| = \frac{6}{5} = 1\frac{1}{5}$.

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 $x$.)

Comments

Post a Comment

Popular posts from this blog

Math Formulas + Facts

Sum of cubes and Difference of Cubes a 3  + b 3  = (a + b) (a 2  − a b + b 2  ) a 3  - b 3  = (a - b) (a 2  + a b + b 2  ) Powers to know 5 7  = 78125 5 6  = 15625 5 5  = 3125 Compound Interest A = P(1+r/n) nt Harmonic Series 1+1/2+1/3+1/4+1/5+1/6...+1/n To estimate sum, take the 2 0  term for 1, then take the next 2 1  terms for 2, then take the next  2 2  terms for 3.... Sums: 2 - up to 1/4 2.5 - up to 1/7 3 - up to 1/11 3.5 - up to 1/19 4 - up to 1/31 4.5 - up to 1/51 5 - up to 1/83 Area of Octogon A = 2(1+√2) a 2 Finding the number sides of a regular polygon when you know the interior angle in degrees 180-x° = 360/n Finding total number of diagonals in a polygon n(n-3)/2
Read more

Discovered New Theorem while solving Chinese Remainder Theorem

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 ...
Read more

Cyclic Quattrocento (Fall 2021 AMC 12B #24)

Image
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...
Read more