What is the sum of all integer values of n satisfying 1≤n≤100, such that n2−1 is a product of exactly two distinct prime numbers?
I did that one yesterday, assuming that the
n2 - 1 meant n^2 - 1
http://www.jiskha.com/display.cgi?id=1364361069
n
is an integer that satisfies .
1
≤
n
≤
100
To find the sum of all integer values of n satisfying the given condition, we need to determine which values of n satisfy the condition and then sum those values.
Let's break down the problem into smaller steps:
Step 1: Determine the prime factorization of n^2 - 1.
The expression n^2 - 1 can be factored using the difference of squares formula:
n^2 - 1 = (n - 1)(n + 1)
Step 2: Check if the factors (n - 1) and (n + 1) are two distinct prime numbers.
For (n - 1) and (n + 1) to be two distinct prime numbers, we need to evaluate the following conditions:
- Both (n - 1) and (n + 1) should be prime numbers.
- They should not be equal.
Step 3: Iterate through all integer values of n from 1 to 100 and check if the conditions are satisfied for each value of n.
For each value of n from 1 to 100, perform the following steps:
- Check if (n - 1) and (n + 1) are prime numbers.
- Check if (n - 1) and (n + 1) are not equal.
Step 4: Sum the values of n that satisfy the conditions.
For each value of n that satisfies the conditions, add the value of n to the total sum.
Let's go through the steps and find the sum of all integer values of n satisfying the given condition:
Step 1: The prime factorization of n^2 - 1 is (n - 1)(n + 1).
Step 2: Both (n - 1) and (n + 1) need to be prime numbers, and they should not be equal.
Step 3: Iterate through n = 1 to 100 and check the conditions.
Step 4: Sum the values of n that satisfy the conditions.
Using this approach, you can find the sum of the integer values of n that satisfy the given condition.