There are 128 three-digit numbers divisible by 7.
This can be determined by finding the first and last three-digit numbers divisible by 7, and then using the formula for the number of terms in an arithmetic progression.
Understanding the Problem
The question asks for the count of three-digit integers (numbers from 100 to 999) that are perfectly divisible by 7, meaning they leave no remainder when divided by 7.
Solution
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Finding the first three-digit number divisible by 7: The smallest three-digit number is 100. Dividing 100 by 7 gives a quotient of 14 with a remainder of 2. Therefore, the first three-digit number divisible by 7 is 105 (14 x 7 + 2).
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Finding the last three-digit number divisible by 7: The largest three-digit number is 999. Dividing 999 by 7 gives a quotient of 142 with a remainder of 5. Therefore, the largest three-digit number divisible by 7 is 994 (142 x 7).
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Using Arithmetic Progression: The numbers divisible by 7 form an arithmetic progression (AP) with a first term (a) of 105, a common difference (d) of 7, and a last term (l) of 994. The formula to find the number of terms (n) in an AP is:
n = (l - a)/d + 1
Substituting the values, we get:
n = (994 - 105)/7 + 1 = 128
Therefore, there are 128 three-digit numbers divisible by 7. This is confirmed by multiple sources (See References).
Examples
- 105, 112, 119 are the first few three-digit numbers divisible by 7.
- 987, 994 are the last few three-digit numbers divisible by 7.
References
Multiple online resources and forums confirm the answer:
- Byjus.com, Toppr.com, Cuemath.com, and other websites and forums independently verify that there are 128 three-digit numbers divisible by 7. Many show the calculation using arithmetic progression as explained above. One source explicitly states: "Therefore, 128 three-digit numbers are divisible by 7."
