Difference between revisions of "2023 SSMO Relay Round 1 Problems"

Latest revision as of 11:32, 15 September 2025

Compute the remainder when $2022^{2021^{2020^{\dots}}}$ is divided by $2023$ .

Let $T=TNYWR$ . Let $a_0 = 3, a_1 = 1, a_2 = T$ , and let $a_n = a_{n-1} - \frac{a_{n-3}}{8}$ for $n \ge3.$ Find $\sum_{i=0}^\infty a_i.$

Let $T=TNYWR$ . Find the number of solutions to the equation $\sec^{T} (Tx) - \tan^{T}(Tx) = 1$ such $0 \le x \le \pi$

@@ Line 7: / Line 7: @@
 ==Problem 2==
-Let <math>T=TNYWR</math>. Let <math>a_0 = 3, a_1 = 1, a_2 = N</math>, and let <math>a_n = a_{n-1} - \frac{a_{n-3}}{8}</math>. Find <cmath>\sum_{i=0}^\infty a_i.</cmath>
+Let <math>T=TNYWR</math>. Let <math>a_0 = 3, a_1 = 1, a_2 = T</math>, and let <math>a_n = a_{n-1} - \frac{a_{n-3}}{8}</math> for <math>n \ge3.</math> Find <cmath>\sum_{i=0}^\infty a_i.</cmath>
 [[2023 SSMO Relay Round 1 Problems/Problem 2|Solution]]
@@ Line 14: / Line 14: @@
 Let <math>T=TNYWR</math>. Find the number of solutions to the equation
-<cmath>\sec^{N} (Nx) - \tan^{N}(Nx) = 1</cmath>
+<cmath>\sec^{T} (Tx) - \tan^{T}(Tx) = 1</cmath>
 such <math>0 \le x \le \pi</math>
 [[2023 SSMO Relay Round 1 Problems/Problem 3|Solution]]