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
To find the last two digits of t3, we need to calculate 3^3^3^3.
First, let's find the value of t1, t2, and t3:
t0 = 3
t1 = 3^3 = 27
t2 = 3^3^3 = 3^27
Now, let's calculate 3^27:
3^27 = 7625597484987 (using a calculator)
To find the last two digits, we take the remainder when dividing by 100:
7625597484987 % 100 = 87
Therefore, the last two digits of t3 = 3^3^3^3 are 87.
To show that the last 10 digits of tk are the same for all k ≥ 10, we need to prove that the remainder when dividing tk by 10^10 is the same for all k ≥ 10.
We can use mathematical induction to prove this.
Base case (k = 10):
t10 = 3^10 = 59049
59049 % 10^10 = 59049, so the remainder is 59049.
Inductive step:
Assume that tk % 10^10 = 59049 for some k ≥ 10.
We will prove that (tk+1) % 10^10 = 59049.
tk+1 = 3(tk) = 3(59049)
Let's calculate 3(59049):
3(59049) = 177147
Now, let's divide 177147 by 10^10:
177147 % 10^10 = 177147, so the remainder is 177147.
Since the remainder when dividing tk+1 by 10^10 is 177147, which is the same as the remainder when dividing tk by 10^10, we have shown that the last 10 digits of tk are the same for all k ≥ 10.
Therefore, the last 10 digits of tk are the same for all k ≥ 10.
To find the last two digits of t3 = 3^3^3^3, we can start by examining the powers of 3 modulo 100 (meaning we consider only the last two digits of each term).
Let's first compute the first few powers of 3 modulo 100:
3^1 ≡ 3 (mod 100)
3^2 ≡ 9 (mod 100)
3^3 ≡ 27 (mod 100)
3^4 ≡ 81 (mod 100)
3^5 ≡ 43 (mod 100)
3^6 ≡ 29 (mod 100)
3^7 ≡ 87 (mod 100)
3^8 ≡ 61 (mod 100)
3^9 ≡ 83 (mod 100)
3^10 ≡ 49 (mod 100)
Notice that the powers of 3 modulo 100 repeat after 20 terms. This is known as the period of the modulo operation.
Now, we can compute the exponent in t3 = 3^3^3^3 step by step:
t0 = 3
t1 = 3^3 = 27
t2 = 3^3^3 ≡ 3^27 (mod 100)
Since 27 ≡ 7 (mod 20), we can substitute t2 as:
t2 ≡ 3^7 ≡ 87 (mod 100)
Now, we can calculate t3 = 3^3^3^3 as:
t3 = 3^(3^(3^3)) ≡ 3^(3^27) = 3^t2 (mod 100)
Since t2 ≡ 87 (mod 100), we substitute it into the equation:
t3 ≡ 3^87 (mod 100)
Again, we find the exponent modulo 20:
87 ≡ 7 (mod 20), so we can rewrite t3 as:
t3 ≡ 3^7 (mod 100)
Finally, we can calculate:
t3 ≡ 3^7 ≡ 87 (mod 100)
Therefore, the last two digits of t3 = 3^3^3^3 are 87.
To show that the last 10 digits of tk are the same for all k ≥ 10, we can use the concept of modular arithmetic and the observation that the powers of 3 modulo 100 repeat every 20 terms.
Let's consider k = 10:
t10 ≡ 3^49 (mod 100)
Since 49 ≡ 9 (mod 20), we can substitute t10 as:
t10 ≡ 3^9 (mod 100)
Now, let's consider k = 11:
t11 ≡ 3^87 (mod 100)
Again, since 87 ≡ 7 (mod 20), we can substitute t11 as:
t11 ≡ 3^7 (mod 100)
By comparing these two equations, we can see that the last 10 digits are the same:
t10 ≡ 3^9 ≡ t11 ≡ 3^7 (mod 100)
This pattern continues for all k ≥ 10. The last 10 digits of tk remain the same because the exponents modulo 20 (which determine the repeating pattern) are the same for all values of k ≥ 10.
does your 3^3^3 mean:
1. (3^3)^3 which would be 27^3 = 19683
or
2. 3^(3^3) ? which would be 3^27 = 7.625... x 10^12
e.g. (9^9)^9 = (387,420,489)^9 or 1.966.. x 10^77 , quite large , but ...
9^(9^9) = 9^(387,420,489) , which is unimaginably larger.
It's the 1.
3^3^3 = 19683