Consider the sequence t0 = 3, t1 = 3^3, t2 = 3^3^3 , t3 = 3^3^3^3,….. , defined by t0 = 3 and tn+1 = 3tn for n ≥ 0.
What are the last 2 digits of t3 = 3^3^3^3 ?
Show that the last 10 digits of tk are the same for all k ≥ 10
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To find the last 2 digits of t3, we will need to calculate the value of t3 first.
Given that t0 = 3 and tn+1 = 3tn, we can calculate the value of t1, t2, and t3 step by step.
t0 = 3
t1 = 3 * t0 = 3 * 3 = 9
t2 = 3 * t1 = 3 * 9 = 27
t3 = 3 * t2 = 3 * 27 = 81
Now, we need to find the last 2 digits of t3. The last 2 digits refer to the remainder obtained when t3 is divided by 100.
t3 mod 100 = 81 mod 100 = 81
Therefore, the last 2 digits of t3 are 81.
To show that the last 10 digits of tk are the same for all k ≥ 10, we need to find a pattern in the values of tk.
Let's calculate the values of t4, t5, and t6 to see if we can observe any patterns:
t4 = 3 * t3 = 3 * 81 = 243
t5 = 3 * t4 = 3 * 243 = 729
t6 = 3 * t5 = 3 * 729 = 2187
From the calculations, we can see that the last digit of each tk is always 3.
Additionally, we can observe that t4 = 243, t5 = 729, and t6 = 2187 all have the same last 3 digits.
By multiplying any number by 3, the last digit remains the same (3).
Therefore, for k ≥ 10, the last digit of tk will always be 3.
Furthermore, we can deduce that for k ≥ 10, tk will have the same last 10 digits as t10, t11, t12, and so on. This is because multiplying by 3 maintains the same last digit, and the last 10 digits are repeated.
Hence, the last 10 digits of tk are the same for all k ≥ 10.