Answer :
Sure, let's solve the problem step by step.
### Part a: Check if 101 and 103 are prime numbers.
To verify that 101 and 103 are prime, we need to check that they are not divisible by any prime numbers up to the square root of each number. The square root of 101 is approximately 10.05, and the square root of 103 is approximately 10.14. Therefore, we need to check divisibility by the prime numbers 2, 3, 5, and 7.
- 101:
- It is not divisible by 2 (it's an odd number).
- The sum of the digits (1 + 0 + 1 = 2) is not divisible by 3.
- It does not end in 0 or 5, so it's not divisible by 5.
- Dividing 101 by 7 gives a non-integer result (approximately 14.4286), indicating it's not divisible by 7.
Since 101 is not divisible by any prime numbers up to its square root, 101 is a prime number.
- 103:
- It is not divisible by 2 (it's an odd number).
- The sum of the digits (1 + 0 + 3 = 4) is not divisible by 3.
- It does not end in 0 or 5, so it's not divisible by 5.
- Dividing 103 by 7 gives a non-integer result (approximately 14.7143), indicating it's not divisible by 7.
Since 103 is also not divisible by any prime numbers up to its square root, 103 is a prime number.
### Part b: Consider [tex]\( P \)[/tex] and [tex]\( M = P + 103 \)[/tex]
Let [tex]\( P \)[/tex] be the product of all the prime numbers from 2 through 101. We are tasked with considering the number [tex]\( M = P + 103 \)[/tex].
- Since [tex]\( P \)[/tex] is a product of all prime numbers up to 101, [tex]\( P \)[/tex] is divisible by each of those primes.
- Therefore, when you add 103 to [tex]\( P \)[/tex], the resulting number [tex]\( M \)[/tex] is not divisible by any of those primes. This is because if you divide [tex]\( P \)[/tex] by any prime from 2 to 101, the remainder is zero (since [tex]\( P \)[/tex] is a multiple of those primes). But when you add 103, the remainder becomes 103, which cannot be zero for any of those primes.
Since [tex]\( M = P + 103 \)[/tex] isn't divisible by any of these primes, it means no prime from 2 through 103 divides [tex]\( M \)[/tex].
### Part c: Existence of Another Prime Greater than 103
The previous step establishes that [tex]\( M \)[/tex] (which equals the product of all primes up to 101, plus 103) is not divisible by any prime up to 103. If there were no primes greater than 103, then [tex]\( M \)[/tex] should be divisible by at least one prime from this set, which contradicts what we found — none of the primes from 2 through 103 divides [tex]\( M \)[/tex].
Therefore, there must be a prime number greater than 103, confirming the existence of another prime beyond the given range.
This completes the detailed solution to the problem based on the information provided.
### Part a: Check if 101 and 103 are prime numbers.
To verify that 101 and 103 are prime, we need to check that they are not divisible by any prime numbers up to the square root of each number. The square root of 101 is approximately 10.05, and the square root of 103 is approximately 10.14. Therefore, we need to check divisibility by the prime numbers 2, 3, 5, and 7.
- 101:
- It is not divisible by 2 (it's an odd number).
- The sum of the digits (1 + 0 + 1 = 2) is not divisible by 3.
- It does not end in 0 or 5, so it's not divisible by 5.
- Dividing 101 by 7 gives a non-integer result (approximately 14.4286), indicating it's not divisible by 7.
Since 101 is not divisible by any prime numbers up to its square root, 101 is a prime number.
- 103:
- It is not divisible by 2 (it's an odd number).
- The sum of the digits (1 + 0 + 3 = 4) is not divisible by 3.
- It does not end in 0 or 5, so it's not divisible by 5.
- Dividing 103 by 7 gives a non-integer result (approximately 14.7143), indicating it's not divisible by 7.
Since 103 is also not divisible by any prime numbers up to its square root, 103 is a prime number.
### Part b: Consider [tex]\( P \)[/tex] and [tex]\( M = P + 103 \)[/tex]
Let [tex]\( P \)[/tex] be the product of all the prime numbers from 2 through 101. We are tasked with considering the number [tex]\( M = P + 103 \)[/tex].
- Since [tex]\( P \)[/tex] is a product of all prime numbers up to 101, [tex]\( P \)[/tex] is divisible by each of those primes.
- Therefore, when you add 103 to [tex]\( P \)[/tex], the resulting number [tex]\( M \)[/tex] is not divisible by any of those primes. This is because if you divide [tex]\( P \)[/tex] by any prime from 2 to 101, the remainder is zero (since [tex]\( P \)[/tex] is a multiple of those primes). But when you add 103, the remainder becomes 103, which cannot be zero for any of those primes.
Since [tex]\( M = P + 103 \)[/tex] isn't divisible by any of these primes, it means no prime from 2 through 103 divides [tex]\( M \)[/tex].
### Part c: Existence of Another Prime Greater than 103
The previous step establishes that [tex]\( M \)[/tex] (which equals the product of all primes up to 101, plus 103) is not divisible by any prime up to 103. If there were no primes greater than 103, then [tex]\( M \)[/tex] should be divisible by at least one prime from this set, which contradicts what we found — none of the primes from 2 through 103 divides [tex]\( M \)[/tex].
Therefore, there must be a prime number greater than 103, confirming the existence of another prime beyond the given range.
This completes the detailed solution to the problem based on the information provided.
