Factors of \(\displaystyle 36\) are:

\(\displaystyle 1,2,3,4,6,9,12, 18\ and\ 36\). So it has \(\displaystyle 9\) factors, which is less than \(\displaystyle 10\).

Factors of \(\displaystyle 125\) are:

\(\displaystyle 1,5,25\ and\ 125\) and their summation is:

\(\displaystyle 1+5+25+125=156\)

and the factors of \(\displaystyle 80\) are:

\(\displaystyle 1,2,4,5,8,10,16,20,40\ and\ 80\) and their summation is:

\(\displaystyle 1+2+4+5+8+10+16+20+40+80=186\)

So \(\displaystyle (b)\) is greater than \(\displaystyle (a)\).

Factors of \(\displaystyle 100\) are \(\displaystyle 1,2,4,5,10,20,25,50\ and\ 100\). So we should compare the mean and the median of the following set of numbers:

\(\displaystyle \left \{ 1,2,4,5,10,20,25,50,100 \right \}\)

The mean of a set of data is given by the sum of the data, divided by the total number of values in the set:

\(\displaystyle Mean=\frac{1+2+4+5+10+20+25+50+100}{9}=\frac{217}{9}\approx24\)

The median is the middle value of a set of data containing an odd number of values which is \(\displaystyle 10\) in this problem. So the mean is greater than the median.

Factors of \(\displaystyle 8\) are: \(\displaystyle 1,2,4\ and \8\). So the product of the factors of \(\displaystyle 8\) are:

\(\displaystyle 1\times 2\times 4\times 8=64\)

The median is the middle value of a set of data containing an odd number of values, which is \(\displaystyle 48\) in this problem. So \(\displaystyle (a)\) is greater than \(\displaystyle (b)\).

 Factors of \(\displaystyle 60\) are \(\displaystyle 1,2,3,4,5,6,10,12,15,20,30\ and\ 60\)

\(\displaystyle \Rightarrow Sum \ of\ the \ factors=1+2+3+4+5+6+10+12+15+20+30+60=168\)

Factors of \(\displaystyle 14\) are \(\displaystyle 1,2,7\ and\14\)

\(\displaystyle \Rightarrow Product \ of\ the\ factors=1\times 2\times 7\times 14=196\)

So \(\displaystyle (b)\) is greater thaan \(\displaystyle (a)\).

If \(\displaystyle x+ y\) is even, then \(\displaystyle x\) and \(\displaystyle y\) are either both even or both odd. The difference of two even numbers is even, and so is the difference of two odd numbers, so \(\displaystyle x - y\) must be even. Let's check the other choices, however:

\(\displaystyle x - y\) must be even, so \(\displaystyle x - y + 1\) must be odd.

\(\displaystyle xy\) is either the product of two even numbers, which is even, or the product of two odd numbers, which is odd. Therefore, \(\displaystyle xy\) is of indeterminate sign. Similarly, \(\displaystyle xy + 1\) is as well.

\(\displaystyle 2y\) is even for any integer \(\displaystyle y\), so \(\displaystyle x + 2 y\) takes the same sign as \(\displaystyle x\); but this is not given to us.

\(\displaystyle x - y\) is the correct choice.

Leaving out 28 itself, the factors of 28 are \(\displaystyle \left \{ 1, 2, 4, 7, 14\right \}\).  The sum of all of these factors is \(\displaystyle 1 + 2 + 4 + 7 + 14 = 28\), making the quantities equal.

An integer divided by 4 yields one of four remainders - 0, 1, 2, or 3. However, if the integer is even, the remainder must also be even, so it must be 0 or 2. Either way, 3 is greater, so (A) is the correct choice.