What are the last three digits of the greatest common divisor of 2^(2^100−1)−1 and 2(2^105−1)−1?
To find the last three digits of the greatest common divisor (GCD) of the given numbers, we can use modular arithmetic. Specifically, we can apply the concept of modular exponentiation.
Let's break down the problem step by step:
Step 1: Find the values of 2^100 and 2^105 modulo 1000.
- Since we are only interested in the last three digits, we can take the numbers modulo 1000.
- To calculate the modulo of a power of 2, we can use the fact that the last three digits of powers of 2 repeat every 500.
2^100 % 1000 = 2^100 % 500 = 376
2^105 % 1000 = 2^105 % 500 = 376
Step 2: Calculate 2^(2^100 - 1) - 1 and 2(2^105 - 1) - 1 modulo 1000.
- We subtract 1 from the exponents in both numbers to get their respective remainders modulo 1000.
2^(2^100 - 1) - 1 % 1000 = 2^(376 - 1) - 1 % 1000 = 2^375 - 1 % 1000
2(2^105 - 1) - 1 % 1000 = 2(376 - 1) - 1 % 1000 = 2^375 - 1 % 1000
Step 3: Find the GCD of the two numbers calculated in Step 2.
- We can now calculate the GCD of the two numbers we obtained in Step 2.
GCD(2^375 - 1 % 1000, 2^375 - 1 % 1000)
Step 4: Determine the last three digits of the GCD.
- Finally, we need to find the last three digits of the GCD obtained in Step 3.
By evaluating the expression GCD(2^375 - 1 % 1000, 2^375 - 1 % 1000), we get the last three digits of the GCD, which are **675**.
Therefore, the last three digits of the GCD of 2^(2^100−1)−1 and 2(2^105−1)−1 are 675.