There are 199 natural numbers less than 1000 that are divisible by 5.
This can be easily calculated by dividing 1000 by 5 and subtracting 1 to exclude 1000 itself. The calculation is: 1000 / 5 = 200, then 200 - 1 = 199.
The sequence of numbers divisible by 5 less than 1000 forms an arithmetic progression (AP): 5, 10, 15, ..., 995. As shown in several references, this can be solved using the formula for the nth term of an AP, or simply by integer division.
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Method 1 (Integer division): The number of multiples of 5 less than 1000 is simply ⌊1000/5⌋ = 200. However, since we want numbers less than 1000, we subtract 1 to exclude 1000 itself, resulting in 199.
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Method 2 (Arithmetic Progression): We have an arithmetic progression with first term a = 5, common difference d = 5, and last term l = 995. The number of terms (n) can be found using the formula l = a + (n-1)d. Solving for n gives n = (l - a)/d + 1 = (995 - 5)/5 + 1 = 199.
Several provided references corroborate this result, although some offer slightly different approaches or focus on related problems involving divisibility by other numbers (e.g., 7, 2, 3). The core concept remains the same: integer division provides the most efficient solution.
