There is no need to be high

The given expression can be written as

$(2^{2} + 1)*2^{10} + (2^{2} - 1)*2^{8} + (2^{4} - 1)*2{4} + (2^{2} - 1)*2^{2} =$

$(2^{12} + 2^{10}) + (2^{10} - 2^{8}) + ((2^{8} - 2^{4}) + (2^{4} - 2^{2}) =$

$2^{12} + 2*2^{10} - 2^{2} = 2^{12} + 2^{11} - 2^{2}.$

Now note that $2^{0} + 2^{1} + 2^{2} + .... + 2^{9} + 2^{10} = 2^{11} - 1,$

and so $2^{11} - 2^{2} = 2^{11} - 1 - 2^{0} - 2^{1} = 2^{2} + 2^{3} + .... + 2^{9} + 2^{10}.$

Thus the binary equivalent of $2^{12} + 2^{11} - 2^{2}$ is $1011111111100_{2},$

which has $\boxed{10}$ 1's.

Work out the denary answer first:

$5 \times 1024 + 3 \times 256 + 15 \times 16 + 3 \times 4 = 5120 + 768 + 240 + 12 =6140$

To convert this to binary, subtract powers of 2 until you get 0. Start with the biggest power of 2 that is smaller than 6140, and note down each time you subtract a power of 2. The total number of times you subtract a power of 2 is the solution, as this is how many ones are in the $2^{0}$ or $2^{1}$ or $2^{2}$ ... columns.

The first number to subtract is $2^{12}$ , which is $4096$ . $6140 - 4096 = 2044$ Next, it is $2^{10}$ as $2^{11}$ ( $2048$ ) is bigger than 2044. Next, it is $2^{9}$ , then $2^{8}$ , $2^{7}$ , $2^{6}$ , $2^{5}$ , $2^{4}$ , $2^{3}$ , and finally $2^{2}$ .

This is a total of 10 numbers subtracted, so there are 10 of those numbers in 6140, hence there are $\boxed{10}$ ones in the binary equivalent of 6140.

multiplying in binary system is like the decimal system, we used to, when you counter zeros, it doesn't change then multiply the non-zero numbers. ex, in decimal,

$2 * 100 = 200$ ,

in binary, $2*100 = 1000$ .

we just multiply the nonzero number but in the binary form. Note that in binary $(2^{x})_{10} = (10^{x})_2$

so, $(5 * 2^{10} )_{10} = (5)_{10} * ( 10^{10} )_{2} = (101)_{2} * (10^{10})_{2}$

similarly,

$(3 * 2^{8} )_{10} = (11)_{2} * (10^{8})_{2}$

$(15 * 2^4)_{10} = (1111)_2 * (10^4)_2$

$(3 * 2^2 ) _{10} = (11)_2 * (10^2)_ 2$

Now, we can easily see that the number : $(11)_2 * (10^2)_ 2 + (1111)_2 * (10^4)_2 + (11)_{2} * (10^{8})_{2} + (101)_{2} * (10^{10})_{2} =$

$1011111111100_2$

and it has $10$ ones