As perfect as a square! 4

Here's my approach.

Let's break down each number into it's respective prime factors:

$10 = 2 \times 5$ $20 = 2^{2} \times 5$ $30 = 2 \times 3 \times 5$

etc...

I don't think this is tedious since it is relatively straightforward to decompose multiples of $10$ into prime factors.

If you then multiply these numbers you will get $2^{16} \times 3^{4} \times 5^{10} \times 7$ . From there I must definitely remove $70$ since there is no other power of $7$ to make a perfect square.

Also by removing $70$ we result in the sum $2^{15} \times 3^{4} \times 5^{9}$ . Notice that if all the exponents are even then it is a perfect square. Therefore we can simply remove $10, 40$ or $90$ which makes the product of the remaining numbers which is now a perfect square (eg. if we remove $10$ then the product will become $2^{14} \times 3^{4} \times 5^{8}$ )

Therefore the answer is $2$ .

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This is the $4$ th problem I made relating to square numbers. Here are the others:

As perfect as a square! 3

As perfect as a square! 2 ⭐

As perfect as a square! 1