ExplanationThis is a combinatorics problem. Make four “slots” (since the numbers are all four-digit numbers), and determine how many possibilities there are for each slot:
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Since the number must begin with 2 or 3, there are two possibilities for the first slot. Because the ones digit must be prime and there are only four prime one-digit numbers (2, 3, 5, and 7), there are four possibilities for the last slot:
2........... 4
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The other slots have no restrictions, so put 10 in them, since there are ten digits from 0–9:
2 10 10 4
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Multiply to get 800.
Alternatively, figure out the pattern and add up the number of qualifying four-digit integers. In the first ten numbers, 2000–2009, there are exactly four numbers that have a prime units digit: 2002, 2003, 2005, and 2007.
The pattern then repeats in the next group of ten numbers, 2010–2020, and so on. In any group of ten numbers, then, four qualify. Between 2,000 and 3,999 there are \(3,999 – 2,000 + 1 = 2,000\) numbers, or \(\frac{2000}{10}= 200\) groups of ten numbers, so there are a total of \(400 \times 2 = 800\) numbers that have a prime units digit.
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