What are 5 different equation that have x=5 as a solution.
1 answer:
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Answer:
More than 50
Step-by-step explanation:
To solve, we need to first see that the function is h(n). Picking main points from the question statement:
- h(n) is the product of all even integers (From 2 to n)
- p is the smallest factor of h(100)+1
- h(100)+1 , here n=100
From here, we can write h(100) as:
h(100) =
h(100) = =
so,
h(100)+1 =
Now two numbers,
h(100) and h(100)+1 are consecutive integers and since they are consecutive so they are co-prime. Hence they only have common factor of 1. Example, 13 and 14 have only common factor of 1
As h(100) has all prime numbers from 1 to 50 and according to above statement h(100)+1 won't have any prime factor from 1 to 50, so the smallest prime factor p is greater than 50.
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