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ollegr [7]
3 years ago
10

Sum of all natural numbers from100to300 which are divisible by 4and 5 is​

Mathematics
1 answer:
My name is Ann [436]3 years ago
4 0

If x is a number that is both divisible by 4 and 5, then

\begin{cases}x\equiv0\pmod4\\x\equiv0\pmod5\end{cases}

4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that x=20n is a solution to the system, where n is any integer. Simply put, any multiple of 20 fits the bill.

Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have 20n=100 when n=5, so the sum we want to compute is

\displaystyle\sum_{n=5}^{15}20n=\boxed{2200}

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