\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
The goal of this exercise is to determine how many zeros are at the end of the number \(1\,000!\) (1,000 factorial).
Recall: \(1\,000! = 1 \times 2 \times 3 \times \dots \times 1\,000\).
  1. Show that there exist integers \(p\) and \(q\) (\(p > 1\) and \(q > 1\)) and an integer \(N\) coprime to \(10\) such that: $$1\,000! = 2^p \times 5^q \times N$$
    1. Among the integers from \(1\) to \(1\,000\), how many are divisible by \(5\)? By \(5^2\)? By \(5^3\)? By \(5^4\)?
    2. Deduce that \(q = 249\).
  3. Show that \(p > q\) and explain why \(q\) is the number of trailing zeros.

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