Problem Solving

find the sum of all even numbers from 10 to 300, excluding those which are multiples of 8.

A 17014
B 2345
C 4562
D 9128
E none of these

vaibhav101 wrote:find the sum of all even numbers from 10 to 300, excluding those which are multiples of 8.

A 17014
B 2345
C 4562
D 9128
E none of these

Hello vaibhav101.

I would solve it as follows: $$10+12+14+\cdots+296+298+300=2\left(5+6+7+8+\cdots+148+149+150\right)$$ $$2\left(\left(1+2+3+4+5+6+7+8+\cdots+148+149+150\right)-\left(1+2+3+4\right)\right)$$ Now, the sum from 1 to 150 is equal to $$1+2+3+\cdots+150=\sum_{i=1}^{150}i=\frac{150\cdot\left(150+1\right)}{2}=75\cdot151=11325.$$ Therefore, we have $$2\left(\sum_{i=1}^{150}i\ \ \ \ -\ \left(1+2+3+4\right)\right)=2\left(11325-\left(10\right)\right)=2\left(11315\right)=22630.$$ But, this last sum includes the multiples of 8, so we have to subtract them.

First, we have to identify which are the multiples of 8:

16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, . . . , 288, 296.

Now, we have to calculate the sum of all the numbers above $$16+24+32+40+48+56+\cdots+288+296=8\left(2+3+4+5+6+7+\cdots+36+37\right)$$ $$8\left(\left(1+2+3+4+5+6+7+\cdots+36+37\right)-1\right)=8\left(\sum_{i=1}^{37}i\ \ \ \ -1\right)=8\left(\frac{37\cdot\left(37+1\right)}{2}\ \ \ -1\right)$$ $$=8\left(\frac{37\cdot38}{2}-1\right)=8\left(37\cdot19-1\right)=8\left(703-1\right)=8\cdot702=5616.$$ Hence, the sum of all even numbers from 10 to 300, excluding those which are multiples of 8 is equal to $$22630-5616=17014.$$ This tells us that the correct answer is the option A.

I hope it is clear enough.

Regards.