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.)

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