JAPAN || A Nice Olympiads Trick

[@RealQinnMalloryu4]

3^3^n3^4 3^4^n^3^3 1^1^n^1^2^2 1^2^2^n+1^3 n^1^1 1^2^n+3^1 2^n+3 (n ➖ 3n+2).

[@jackkalver4644]

Here’s a better solution: 7^(n+6)=6^(n+7) 7^(n+6)=6^(n+6)6 (7/6)^(n+6)=6 n+6=log(6, 7/6) n=log(6, 7/6)-6

[@nemesis2022pf]

6^(n+7)=7^(n+6) ln(6^(n+7))=ln(7^(n+6)) (n+7)ln6=(n+6)ln7 n(ln6-ln7)=6ln7-7ln6 n=(6ln7-7ln6)/(ln6-ln7)

[@ruschein]

When you calculate the answer with a calculator you don't get a "more precise answer!" In fact, you get a less precise answer! Nonetheless it's interesting what the numerical, approximate answer is but I am too lazy to do it myself.