No-Perfect-Square-Factors Sequence

The answer of the problem is equal to $\left(\displaystyle \sum_{k=2}^{100}\sqrt{k}\right)-x$ , with $x$ being the sum of the roots of the multiples of perfect square numbers from $2$ to $100$ .

Let's list all multiples of perfect square numbers ranging from $2$ to $100$ :

$\color{#D61F06} 4,8,12,16,20,...,100$ $\implies$ (multiples of $4$ )

$\color{#EC7300} 9,18,27,45,54,...,99$ $\implies$ (multiples of $9$ excluding $\color{#D61F06} 36$ and $\color{#D61F06} 72$ since they're already multiples of $4$ )

$\color{#3D99F6} 25,50,75$ $\implies$ (multiples of 25)

$\color{#20A900} 49,98$ $\implies$ (multiples of $49$ )

The sum of the roots of the numbers in the list above will be $x$ . In the subtraction operation mentioned above, the difference will be the sum of the roots of the numbers from $2$ to $100$ that doesn't have any perfect square factors which is the answer of the problem.

$x=\color{#D61F06}\displaystyle \sum_{k=1}^{25}\sqrt{4k} \color{#333333}+ \color{#EC7300} \displaystyle\sum_{k=1}^{3}\sqrt{9k} \color{#333333}+ \color{#EC7300} \displaystyle \sum_{k=5}^{7}\sqrt{9k} \color{#333333}+ \color{#EC7300} \displaystyle \sum_{k=9}^{11}\sqrt{9k} \color{#333333}+ \color{#3D99F6} \displaystyle \sum_{k=1}^{3}\sqrt{25k} \color{#333333}+ \color{#20A900} \displaystyle \sum_{k=1}^{2}\sqrt{49k}$

Calculating this, it results to $x\approx 271.767805$ . As for $\displaystyle \sum_{k=2}^{100}\sqrt{k}$ , the sum is $\approx 670.462947$ .

$\left(\displaystyle \sum_{k=2}^{100}\sqrt{k}\right)-x\approx 670.462947-271.767805\approx \boxed{398.695}$

Simple Python code to do this:

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required_list = []

for num in range(2, 98):
    count = 0

    for div in range(2, num):
        if num % div**2 == 0:
            break
    else:
        required_list.append(num**0.5)

print(sum(required_list))

Similar solution as @Kaizen Cyrus 's

The sum is the sum of all square-roots from $2$ to $99$ subtracts the sum of roots of multiples of $4$ , $9$ , $25$ , and $49$ , and add back the double count of the multiples of $36 (= 4 \times 9)$ from $2$ to $99$ . In formula we have:

$\begin{aligned} S & = \sum_{k=2}^{99} \sqrt k - 2 \sum_{k=1}^{\left \lfloor \frac {99}4 \right \rfloor} \sqrt k - 3 \sum_{k=1}^{\left \lfloor \frac {99}9 \right \rfloor} \sqrt k - 5 \sum_{k=1}^{\left \lfloor \frac {99}{25} \right \rfloor} \sqrt k - 7 \sum_{k=1}^{\left \lfloor \frac {99}{49} \right \rfloor} \sqrt k + 6 \sum_{k=1}^{\left \lfloor \frac {99}{36} \right \rfloor} \sqrt k \\ & = \sum_{k=2}^{99} \sqrt k - 2 \sum_{k=1}^{24} \sqrt k - 3 \sum_{k=1}^{11} \sqrt k - 5 \sum_{k=1}^3 \sqrt k - 7 \sum_{k=1}^2 \sqrt k + 6 \sum_{k=1}^2 \sqrt k \\ & \approx 660.4629471 - 2(80.63378028) - 3(25.78490298) - 5(4.14626437) - 2.414213562 \\ & \approx \boxed{398.695} \end{aligned}$