For $1$st minimum element there is maximum $0$ comparison

For $2$nd minimum element there is maximum $1$ comparison

For $3$rd minimum element there is maximum $2$ comparison

For $4$st minimum element there is maximum $3$ comparison

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For $65$st minimum element there is maximum $64$ comparison

So, total number of comparisons $0+1+2+...+64$

$\frac{64\times 65}{2}=2080$

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Here we operating on min heap and searching for minimum number of comparison. Now, we can compare when atleast $2$ elements in the heap.So, finding minimum element, i.e. at root, we need no comparison

Now, take an example and see for 2nd , 3rd, 4th minimum how many comparison there can be

Now, like this way for $4$ th minimum we need comparison between 4,5,6,8 , So, there must be 3 comparison. Now like this if we go for $5$ th minimum there are 4 comparison

So, there will be total $2080$ comparisons

Note: We just operating like this; When got some minimum, we just add it's child for next comparison and remaining element of previous comparison

Say for $2$ nd minimum, we take child of $1$ st minimum

For $3$ rd minimum , we take remaining element of $2$ nd minimum i.e. $8$ and child of $2$ i.e. $3,6$