What is the least number of elements we have to delete from the set {10, 20, 30, 40, 50, 60, 70, 80, 90} so that the product of the elements remaining in the set is a perfect square?
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1What is your effort? – MANI Dec 09 '19 at 15:30
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You have to remove $50,70,80$. – Rushabh Mehta Dec 09 '19 at 15:33
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@DonThousand I think removing $70$, $90$ is enough? – ab123 Dec 09 '19 at 15:34
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Note that $$\prod S_i=10^{10}*2^8*3^4*5^2*7$$ To make this a square you must remove the 7 of course, so removing the factor of 70 yields:
$$10^{9}*2^8*3^4*5^2$$
So now you have an extra power of ten, so removing a ten along with any perfect square (1,4, or 9) should suffice.
So the answer is 2, removing either (70,10) (70,40) or (70,90)
Gabe
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You obviously have to remove the $70$, as there is no other power of $7$ to make a perfect square.
But now, you also have to remove one among the remaining numbers as there is an odd power of $5$ overall.
Removing $30$ or $60$ will disrupt the already even power of $3$. Power of $2$ in the remaining product (after removing $70$) is odd. Also, an odd power of $5$ remains.
Removing $90$, $10$ or $40$ suffices (none of them disrupts any remaining powers as they are a multiple of $5 \times 2 \times \text{some perfect square}$ )
Hence $2$ is the answer
ab123
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