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Tens Digit of the Power Tower 7, 7^7, ... (t1000) | JEE

JEE Maths question with a full step-by-step solution.

Question
Let t1=7t_1=7, t2=77t_2=7^7, t3=777t_3=7^{7^7}, and so on. The digit at the tens place of t1000t_{1000} is
A88
B00
C66
D44correct
Solution
Step 1: The tens digit depends only on the last two digits. Since 74=24017^4=2401 ends in 0101, the number 74k+r7^{4k+r} ends in the same two digits as 7r7^r; so the last two digits of a power of 77 depend on the exponent modulo 44. Step 2: Now t1000=7 t999t_{1000}=7^{\,t_{999}}, so reduce t999(mod4)t_{999}\pmod 4. As 7≡−1(mod4)7\equiv-1\pmod 4 and t999t_{999} has an odd exponent,
t999≡(−1)odd≡3(mod4).t_{999}\equiv(-1)^{\text{odd}}\equiv 3\pmod 4.
 Step 3: Hence t1000t_{1000} ends like 73=3437^3=343, i.e. last two digits 4343. Tens digit =4.=4. Correct answer: (4)
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