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Sample Question 19:
Step 1: Compute the prime factorization of 100,000. The factorization is \(2^5 \cdot 5^5\).
Step 2: From that factorization, choose two numbers \(2^a5^b\) and \(2^c5^d\) such that \(0 \le a,b,c,d \le 5\) and \((a,b) \neq (c,d)\). The product of these numbers will be \(2^{a+c}5^{b+d}\), with \(0 \le a+c \le 10\) and \(0 \le b+d \le 10\).
Step 3: Note that these conditions are equivalent to choosing a divisor of \(100,000^2\), which equals \(2^{10}5^{10}\). The divisor count of this number is \(121\), calculated as \((10+1)(10+1)\).
Step 4: Some divisors of \(2^{10}5^{10}\) can't be written as the product of two distinct divisors of \(2^5 \cdot 5^5\). These are \(1 = 2^05^0\), \(2^{10}5^{10}\), \(2^{10}\), and \(5^{10}\). Eliminate those, which reduces the count to 117.
Step 5: Every number of form \(2^p5^q\), with \(0 \le p, q \le 10\) and neither \(p, q\) are both \(0\) or \(10\), can be written as the product of two distinct elements in \(S\). Hence, the answer is \(117\), which corresponds to option C.
Let \(S\) be the set of all positive integer divisors of \(100,000.\) How many numbers are the product of two distinct elements of \(S?\)
\(\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121\)
Answer Keys
Question 19: CSolutions
Question 19Step 1: Compute the prime factorization of 100,000. The factorization is \(2^5 \cdot 5^5\).
Step 2: From that factorization, choose two numbers \(2^a5^b\) and \(2^c5^d\) such that \(0 \le a,b,c,d \le 5\) and \((a,b) \neq (c,d)\). The product of these numbers will be \(2^{a+c}5^{b+d}\), with \(0 \le a+c \le 10\) and \(0 \le b+d \le 10\).
Step 3: Note that these conditions are equivalent to choosing a divisor of \(100,000^2\), which equals \(2^{10}5^{10}\). The divisor count of this number is \(121\), calculated as \((10+1)(10+1)\).
Step 4: Some divisors of \(2^{10}5^{10}\) can't be written as the product of two distinct divisors of \(2^5 \cdot 5^5\). These are \(1 = 2^05^0\), \(2^{10}5^{10}\), \(2^{10}\), and \(5^{10}\). Eliminate those, which reduces the count to 117.
Step 5: Every number of form \(2^p5^q\), with \(0 \le p, q \le 10\) and neither \(p, q\) are both \(0\) or \(10\), can be written as the product of two distinct elements in \(S\). Hence, the answer is \(117\), which corresponds to option C.