Let p, q, r and s be positive natural numbers having three exact factors including 1 and the number itself. If q > p and both are two-digit numbers, and r > s and both are one-digit numbers, then the value of the expression \(\frac{p-q-1}{r-s}\) is:
–s – 1
The core idea in this problem involves identifying numbers with precisely three exact factors. A fundamental rule in number theory states that only the squares of prime numbers have exactly three factors. These factors are always 1, the prime number itself, and the square of that prime number.
Let's list the squares of the first few prime numbers to see which ones fit the criteria:
The question specifies that p and q are positive natural numbers possessing exactly three factors. Furthermore, both are two-digit numbers, and it's given that q > p.
From our list above, the two-digit numbers that have exactly three factors are 25 and 49.
Given the condition $q > p$, we assign the values as follows:
Similarly, r and s are positive natural numbers with exactly three factors. They are required to be one-digit numbers, with the condition r > s.
Looking again at our list, the one-digit numbers with exactly three factors are 4 and 9.
Applying the condition $r > s$, we assign the values:
We need to compute the value of the expression $\frac{p-q-1}{r-s}$.
Let's substitute the determined values of p, q, r, and s into the expression:
The expression becomes:
$$ \frac{25 - 49 - 1}{9 - 4} $$First, we calculate the numerator:
$$ 25 - 49 - 1 = -24 - 1 = -25 $$Next, we calculate the denominator:
$$ 9 - 4 = 5 $$Finally, we perform the division:
$$ \frac{-25}{5} = -5 $$Therefore, the value of the expression $\frac{p-q-1}{r-s}$ is -5.
The final step is to compare our calculated value, -5, with the provided options. The options are expressed in terms of $s$, and we found $s=4$. Let's evaluate each option:
Our calculated result of -5 perfectly matches the value obtained from Option 1.
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