New Light Math

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...