Find the sum

Find the sum of all natural numbers from 1 and 100, which are divisible by 2 or 5

Result

s =  3050

Solution:

2,4,6,8,…,100 5,10,15,…,100  s1=2⋅ (1+2+3+4...+50) s1=n12⋅ (a50+a1) s1=502⋅ (100+2)=2550  s2=5(1+3+5+…+19) s2=n22⋅ (b19+b1) s2=102⋅ (95+5)=500  s=s1+s2=2550+500=30502,4,6,8,…,100 \ \\ 5,10,15,…,100 \ \\ \ \\ s_{1}=2 \cdot \ (1+2+3+4 ... +50) \ \\ s_{1}=\dfrac{ n_{1} }{ 2 } \cdot \ (a_{50}+a_{1}) \ \\ s_{1}=\dfrac{ 50 }{ 2 } \cdot \ (100+2)=2550 \ \\ \ \\ s_{2}=5(1 + 3 + 5 + … + 19) \ \\ s_{2}=\dfrac{ n_{2} }{ 2 } \cdot \ (b_{19}+b_{1}) \ \\ s_{2}=\dfrac{ 10 }{ 2 } \cdot \ (95+5)=500 \ \\ \ \\ s=s_{1}+s_{2}=2550+500=3050

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