AIME II 2025 (Problem 4)The product ∏k=463logk(5k2−1)logk+1(5k2−4)\displaystyle \prod_{k=4}^{63}\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}k=4∏63logk+1(5k2−4)logk(5k2−1) is equal to mn\frac{m}{n}nm, where mmm and nnn are relatively prime positive integers. Find m+nm+nm+n.Related TopicsToolkit 5 — Telescoping seriesToolkit 10 — Difference of squaresToolkit 48 — Logarithms: Definition and PropertiesHints (5)Hint 1logk(5k2−1)logk+1(5k2−4)=(k2−1)logk5(k2−4)logk+15=(k−1)(k+1)logk5(k−2)(k+2)logk+15\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}=\frac{(k^2-1)\log_k5}{(k^2-4)\log_{k+1}5}=\frac{(k-1)(k+1)\log_k5}{(k-2)(k+2)\log_{k+1}5}logk+1(5k2−4)logk(5k2−1)=(k2−4)logk+15(k2−1)logk5=(k−2)(k+2)logk+15(k−1)(k+1)logk5Hint 2∏k=463logk(5k2−1)logk+1(5k2−4)\prod_{k=4}^{63}\frac{\log_k(5^{k^2-1})}{\log_{k+1}(5^{k^2-4})}k=4∏63logk+1(5k2−4)logk(5k2−1)=3⋅5log452⋅6log55⋅4⋅6log553⋅7log65⋅5⋅7log654⋅8log75⋯62⋅64log63561⋅65log645=\frac{3\cdot5\log_4 5}{2\cdot6\log_5 5}\cdot\frac{4\cdot6\log_5 5}{3\cdot7\log_6 5}\cdot\frac{5\cdot7\log_6 5}{4\cdot8\log_7 5}\cdots\frac{62\cdot64\log_{63}5}{61\cdot65\log_{64}5}=2⋅6log553⋅5log45⋅3⋅7log654⋅6log55⋅4⋅8log755⋅7log65⋯61⋅65log64562⋅64log635Hint 3Rearrange the numbers:3⋅5log452⋅6log55⋅4⋅6log553⋅7log65⋅5⋅7log654⋅8log75⋯62⋅64log63561⋅65log645\frac{3\cdot5\log_4 5}{2\cdot6\log_5 5}\cdot\frac{4\cdot6\log_5 5}{3\cdot7\log_6 5}\cdot\frac{5\cdot7\log_6 5}{4\cdot8\log_7 5}\cdots\frac{62\cdot64\log_{63}5}{61\cdot65\log_{64}5}2⋅6log553⋅5log45⋅3⋅7log654⋅6log55⋅4⋅8log755⋅7log65⋯61⋅65log64562⋅64log635=3⋅4⋅5⋯62 ⋅ 5⋅6⋅7⋯642⋅3⋅4⋯61 ⋅ 6⋅7⋅8⋯65⋅log45log55⋅log55log65⋯log635log645=\frac{3\cdot4\cdot5\cdots62\;\cdot\;5\cdot6\cdot7\cdots64}{2\cdot3\cdot4\cdots61\;\cdot\;6\cdot7\cdot8\cdots65}\cdot\frac{\log_4 5}{\log_5 5}\cdot\frac{\log_5 5}{\log_6 5}\cdots\frac{\log_{63}5}{\log_{64}5}=2⋅3⋅4⋯61⋅6⋅7⋅8⋯653⋅4⋅5⋯62⋅5⋅6⋅7⋯64⋅log55log45⋅log65log55⋯log645log635=62⋅52⋅65⋅log45log645=\frac{62\cdot5}{2\cdot65}\cdot\frac{\log_4 5}{\log_{64}5}=2⋅6562⋅5⋅log645log45Hint 4log645=log435=13log45\log_{64}5=\log_{4^3}5=\frac13\log_4 5log645=log435=31log45Hint 562⋅52⋅65⋅log45log645=3113⋅3=9313\frac{62\cdot5}{2\cdot65}\cdot\frac{\log_4 5}{\log_{64}5}=\frac{31}{13}\cdot3=\frac{93}{13}2⋅6562⋅5⋅log645log45=1331⋅3=1393Final Answer106106106