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Find the sum of all positive integers less than 1000 ending in 3 or 4 or 6 or 9.

 May 9, 2024
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We will tackle the general case first. What is the sum of all positive integers less than 1000 ending in a digit n?

The numbers must be of the form \(10m + n\) where \(0 \leq m \leq 99\).Therefore, the answer will be \(\displaystyle \sum_{m = 0}^{99} (10m + n) = 100n + 49500\).

 

Hence, the answer to your original question will be \(4(49500) + 100(3 + 4 + 6 + 9) = 200200\).

 May 10, 2024

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