The "random" numbers produced by computers aren't purely random. They are actually pseudo-random, meaning that they are produced by mathematical formulas that simulate randomness.The linear congruential generator takes a seed X0 and generates subsequent pseudo-random numbers using the formula:
Xn+1=(aXn+c) mod m
X1 is the first pseudo-random number generated, X2 is the second, and so on. Let R be the 2000th pseudo-random number generated by the linear congruential generator when X0=42, a=25, c=31, and m=2^20. What are the last three digits of R?
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If the question asks for the last digits, you could code and evaluate the recursive formula to get the results.
If you would like a check, you are welcome. The sum of the 3 last digits is 20.
To find the last three digits of R, we need to calculate the value of R using the given formula for the linear congruential generator.
Xn+1 = (aXn + c) mod m
Given:
X0 = 42
a = 25
c = 31
m = 2^20
First, let's calculate X1 using the formula:
X1 = (25 * 42 + 31) mod (2^20)
= (1050 + 31) mod 1048576
= 1081 mod 1048576
= 1081
Now, let's calculate X2 using X1:
X2 = (25 * 1081 + 31) mod (2^20)
= (27025 + 31) mod 1048576
= 27056 mod 1048576
= 27056
We need to repeat this process till we reach the 2000th pseudo-random number. However, calculating each step manually would be time-consuming and not efficient. Instead, we can use a programming language or a calculator that supports modular arithmetic.
Using a programming language like Python, we can write a code to calculate the 2000th pseudo-random number. Here's an example code in Python:
```python
def linear_congruential_generator(X0, a, c, m, n):
Xn = X0
for _ in range(n):
Xn = (a * Xn + c) % m
return Xn
X0 = 42
a = 25
c = 31
m = 2**20
n = 2000
R = linear_congruential_generator(X0, a, c, m, n)
```
After running the code, the value of R will be the 2000th pseudo-random number generated by the linear congruential generator.
Finally, to find the last three digits of R, we can use the modulo operator (%) to obtain the remainder when dividing R by 1000:
last_three_digits = R % 1000
So, the last three digits of R are the value of `last_three_digits`.