AMC 12B 2025 (Problem 9)What is the tens digit of 6666^{6^6}666?(A) 1\text{(A)}\;1(A)1(B) 3\text{(B)}\;3(B)3(C) 5\text{(C)}\;5(C)5(D) 7\text{(D)}\;7(D)7(E) 9\text{(E)}\;9(E)9Solution 1Related TopicsCoreToolkit 116 — Chinese Remainder Theorem (CRT): Combining CongruencesToolkit 51 — Euler's Totient Function φ(n)Toolkit 52 — Euler's TheoremMajorToolkit 47 — Modular Arithmetic: Definition and PropertiesHints (8)Hint 1If we find 666(mod100), then we have the last 2 digits.\text{If we find }6^{6^6}\pmod{100}\text{, then we have the last 2 digits.}If we find 666(mod100), then we have the last 2 digits.Hint 2666≡?(mod100)6^{6^6}\equiv ?\pmod{100}666≡?(mod100)⟺\Longleftrightarrow⟺666≡?(mod4)6^{6^6}\equiv ?\pmod4666≡?(mod4)666≡?(mod25)6^{6^6}\equiv ?\pmod{25}666≡?(mod25)Hint 3666≡0(mod4)6^{6^6}\equiv0\pmod4666≡0(mod4)Hint 4By Toolkit 51: Euler’s Totient Function φ(n)\text{By Toolkit 51: Euler's Totient Function }\varphi(n)By Toolkit 51: Euler’s Totient Function φ(n)φ(25)=25(1−15)=25−5=20\varphi(25)=25\left(1-\frac15\right)=25-5=20φ(25)=25(1−51)=25−5=20Hint 5By Toolkit 52: Euler’s Theorem\text{By Toolkit 52: Euler's Theorem}By Toolkit 52: Euler’s Theoremgcd(6,25)=1⇒6φ(25)=620≡1(mod25)\gcd(6,25)=1\Rightarrow6^{\varphi(25)}=6^{20}\equiv1\pmod{25}gcd(6,25)=1⇒6φ(25)=620≡1(mod25)Hint 6For 666≡?(mod25), we should find 66≡?(mod20).\text{For }6^{6^6}\equiv ?\pmod{25}\text{, we should find }6^6\equiv ?\pmod{20}.For 666≡?(mod25), we should find 66≡?(mod20).66≡363≡(−4)3≡−64≡16(mod20)6^6\equiv36^3\equiv(-4)^3\equiv-64\equiv16\pmod{20}66≡363≡(−4)3≡−64≡16(mod20)Hint 7616≡368≡118≡1214≡(−4)4(mod25)6^{16}\equiv36^8\equiv11^8\equiv121^4\equiv(-4)^4\pmod{25}616≡368≡118≡1214≡(−4)4(mod25)≡44≡256≡6(mod25)\equiv4^4\equiv256\equiv6\pmod{25}≡44≡256≡6(mod25)Hint 8By Hints 3 and 7\text{By Hints 3 and 7}By Hints 3 and 7666≡0(mod4)6^{6^6}\equiv0\pmod4666≡0(mod4)666≡6≡31≡56(mod25)6^{6^6}\equiv6\equiv31\equiv56\pmod{25}666≡6≡31≡56(mod25)⇒666≡56(mod100)\Rightarrow6^{6^6}\equiv56\pmod{100}⇒666≡56(mod100)⇒Ans = Tens digit=5\Rightarrow\text{Ans = Tens digit}=5⇒Ans = Tens digit=5Related Problems (4)AMC 10B/12B 2024 (Problem 18/14)AMC 10B Fall 2021 (Problem 22)AIME I 2026 (Problem 8)AMC 10B/12B 2025 (Problem 8/6)Solution 2Related TopicsCoreToolkit 75 — Try Small ExamplesMajorToolkit 117 — Finding Last DigitsHints (3)Hint 161→066^1\to0661→0662→366^2\to3662→3663→166^3\to1663→1664→966^4\to9664→9665→766^5\to7665→7666→566^6\to5666→5667→366^7\to3667→36Hint 206;36,16,96,76,56;36,16,…,56;…06;36,16,96,76,56;36,16,\ldots,56;\ldots06;36,16,96,76,56;36,16,…,56;…Period Length=5\text{Period Length}=5Period Length=5Hint 3(66−1)÷5⇒Remainder=16−1=0(6^6-1)\div5\Rightarrow\text{Remainder}=1^6-1=0(66−1)÷5⇒Remainder=16−1=0⇒Last 2 digits=56⇒Ans = Tens digit=5\Rightarrow\text{Last 2 digits}=56\Rightarrow\text{Ans = Tens digit}=5⇒Last 2 digits=56⇒Ans = Tens digit=5Related Problems (10)AMC 12A 2023 (Problem 22)AMC 10A 2025 (Problem 19)AMC 10A/12A 2025 (Problem 21/15)AMC 10B/12B 2025 (Problem 2/2)AMC 10B 2025 (Problem 3)AMC 10B/12B 2025 (Problem 17/14)AMC 10B 2025 (Problem 18)AMC 10B/12B 2025 (Problem 23/19)View all related problems →Solution 3Related TopicsCoreToolkit 118 — Solving Modulo Prime Powers (Binomial Expansion Method)Toolkit 116 — Chinese Remainder Theorem (CRT): Combining CongruencesMajorToolkit 47 — Modular Arithmetic: Definition and PropertiesHints (5)Hint 1If we find 666(mod100), then we have the last 2 digits.\text{If we find }6^{6^6}\pmod{100}\text{, then we have the last 2 digits.}If we find 666(mod100), then we have the last 2 digits.Hint 2666≡?(mod100)6^{6^6}\equiv ?\pmod{100}666≡?(mod100)⟺\Longleftrightarrow⟺666≡?(mod4)6^{6^6}\equiv ?\pmod4666≡?(mod4)666≡?(mod25)6^{6^6}\equiv ?\pmod{25}666≡?(mod25)Hint 3666≡0(mod4)6^{6^6}\equiv0\pmod4666≡0(mod4)Hint 4666≡?(mod25)6^{6^6}\equiv ?\pmod{25}666≡?(mod25)666=(5+1)66=(660)50+(661)51+(662)52+⋯6^{6^6}=(5+1)^{6^6}=\binom{6^6}{0}5^0+\binom{6^6}{1}5^1+\binom{6^6}{2}5^2+\cdots666=(5+1)66=(066)50+(166)51+(266)52+⋯≡1+66⋅5≡1+65⋅30≡1+64⋅30(mod25)\equiv1+6^6\cdot5\equiv1+6^5\cdot30\equiv1+6^4\cdot30\pmod{25}≡1+66⋅5≡1+65⋅30≡1+64⋅30(mod25)≡1+64⋅5≡⋯≡1+5≡6(mod25)\equiv1+6^4\cdot5\equiv\cdots\equiv1+5\equiv6\pmod{25}≡1+64⋅5≡⋯≡1+5≡6(mod25)Hint 5By Hints 3 and 4\text{By Hints 3 and 4}By Hints 3 and 4666≡0(mod4)6^{6^6}\equiv0\pmod4666≡0(mod4)666≡6≡31≡56(mod25)6^{6^6}\equiv6\equiv31\equiv56\pmod{25}666≡6≡31≡56(mod25)⇒666≡56(mod100)\Rightarrow6^{6^6}\equiv56\pmod{100}⇒666≡56(mod100)⇒Ans = Tens digit=5\Rightarrow\text{Ans = Tens digit}=5⇒Ans = Tens digit=5Related Problems (3)AMC 10B Fall 2021 (Problem 22)AIME I 2026 (Problem 8)AMC 10B/12B 2025 (Problem 8/6)Final Answer(C) 555