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The function \(f(n)\) is defined for all positive integers \(n\), where \(n > 1\), as the product of \(n\) consecutive positive multiples of \(n\). How many zeros does \(f(100)\) end with?
A. 24
B. 26
C. 143
D. 224
E. 2400
A. 24
B. 26
C. 143
D. 224
E. 2400
12 Jan 2026, 11:22
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Official Solution:
The function \(f(n)\) is defined for all positive integers \(n\), where \(n > 1\), as the product of \(n\) consecutive positive multiples of \(n\). How many zeros does \(f(100)\) end with?
A. 24
B. 26
C. 143
D. 224
E. 2400
Let's break down the question:
According to the problem, \(f(100)\) is the product of 100 consecutive positive multiples of 100. So:
\(f(100) = 100 * 200 * 300 * ... * 10,000 =\)
\(= (100 * 1) * (100 * 2) * (100 * 3) * ... * (100 * 100) =\)
\(= 100^{100} * 100!\)
Next:
\(100^{100} = (10^2)^{100} = 10^{200}\), which ends with 200 zeros.
\(100!\) ends with \(\frac{100}{5} + \frac{100}{25} = 20 + 4 = 24\) zeros.
Thus, the total number of zeros at the end of \(f(100) = 10^{200} *\) (something with 24 zeros at the end) is:
\(200 + 24 = 224\) zeros.
Answer: D
The function \(f(n)\) is defined for all positive integers \(n\), where \(n > 1\), as the product of \(n\) consecutive positive multiples of \(n\). How many zeros does \(f(100)\) end with?
A. 24
B. 26
C. 143
D. 224
E. 2400
Let's break down the question:
According to the problem, \(f(100)\) is the product of 100 consecutive positive multiples of 100. So:
\(f(100) = 100 * 200 * 300 * ... * 10,000 =\)
\(= (100 * 1) * (100 * 2) * (100 * 3) * ... * (100 * 100) =\)
\(= 100^{100} * 100!\)
Next:
\(100^{100} = (10^2)^{100} = 10^{200}\), which ends with 200 zeros.
\(100!\) ends with \(\frac{100}{5} + \frac{100}{25} = 20 + 4 = 24\) zeros.
Thus, the total number of zeros at the end of \(f(100) = 10^{200} *\) (something with 24 zeros at the end) is:
\(200 + 24 = 224\) zeros.
Answer: D
