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 t3-3^3^3^3, we need to calculate the value of t3-3^3^3^3.
Let's first break down the given sequence:
t0 = 3
t1 = 3^3 = 27
t2 = 3^(3^3) = 3^27
t3 = 3^(3^3^3) = 3^(3^(3^3))
To calculate t3, we need to start from the rightmost operation and work our way up.
First, let's calculate 3^3 = 27.
Next, we raise 3 to the power of 27, i.e., 3^27.
Finally, we raise 3 to the power of the result obtained from the previous step, i.e., 3^(3^27).
Calculating such a huge exponentiation manually could be challenging. Instead, we can make use of modular arithmetic, specifically the concept of congruence.
Let's take an example: 3^5.
If we calculate 3^5 directly, we get 243. However, if we are interested in the last two digits only, we can find the same result by taking 3^5 mod 100. Here, mod 100 means we only consider the last two digits of the result.
So, 3^5 mod 100 = 43.
Let's apply this concept to our problem. We will calculate t3 mod 100 to find the last 2 digits.
Step 1: Calculate 3^27 mod 100.
3^27 mod 100 = 3^7 mod 100 * 3^20 mod 100
Here, calculating 3^7 directly would involve large numbers. Instead, we can use the concept of modular arithmetic:
3^7 mod 100 = (3^6 mod 100 * 3 mod 100) mod 100 = (729 mod 100 * 3 mod 100) mod 100 = (29 * 3) mod 100
= 87 mod 100
= 87
Now, let's calculate 3^20 mod 100 using a similar approach:
3^20 mod 100 = (3^10 mod 100)^2 mod 100
Again, calculating 3^10 directly would involve large numbers. Using modular arithmetic, we have:
3^10 mod 100 = (3^5 mod 100)^2 mod 100 = 43^2 mod 100 = 1849 mod 100 = 49
So, 3^20 mod 100 = 49^2 mod 100 = 2401 mod 100 = 1
Now, we have the values: 3^7 mod 100 = 87 and 3^20 mod 100 = 1.
Step 2: Calculate t3 mod 100.
t3 mod 100 = (3^7 mod 100)^1 mod 100
= 87^1 mod 100
= 87
Thus, the last 2 digits of t3 - 3^3^3^3 are 87.
Now, let's prove that the last 10 digits of tk are the same for all k ≥ 10.
We can observe that after t9, every term in the sequence will be in the form of 3^3^3^3... (repeated exponentiation of 3 with the base 3). In other words, tn will always be a huge number raised to a power of 3.
When calculating t10, t11, t12, and so on, we will be raising such huge numbers with massive exponents.
However, if we focus on finding the last 10 digits only, we can ignore all previous digits and only consider the last 10 digits of the number being raised.
For example, let's consider t10:
t10 = 3^t9 = 3^(huge number)
To find the last 10 digits of t10, we only need to focus on the last 10 digits of the huge number being raised.
Similarly, for t11, t12, and all future terms, we will only focus on the last 10 digits of the number being raised.
Since we are repeatedly taking the last 10 digits of the number being raised, it follows that the last 10 digits of all tk (k ≥ 10) will be the same.