The product of a group of nonzero numbers is positive if and only if an even number of these factors is negative. This occurs in each of these scenarios.

For a number to be divisible by 6, it must be both even and a multiple of 3. None of the choices can be eliminated by that first criterion. But to fit the second criterion, the digit sum must be divisible by 3. We need only add the nonzero digits of each choice, remembering to include the leading 1 in the sum:

\(\displaystyle 42: 1 + 4 + 2 = 7\)

\(\displaystyle 52: 1 + 5 + 2 = 8\)

\(\displaystyle 56: 1 + 5 + 6 = 12\)

\(\displaystyle 66: 1 + 6 + 6 = 13\)

\(\displaystyle 76: 1 + 7 + 6 = 14\)

Of these choices, only 56 passes the divisibility test, so it is the correct choice.

All of these expressions are equivalent to the correct quotient. But a number in scientific notation takes the form

\(\displaystyle a \times 10^{n}\) , where \(\displaystyle 1 \leq |a| < 10\)

Only \(\displaystyle 6.25 \times 10^{10}\) takes this form.

One way to do this:

Divide 117 by 8. The remainder will be the last digit.

\(\displaystyle 117 \div 8 = 14 \textrm{ R }5\)

Now divide the quotient by 8. This remainder will be the second-to-last digit.

\(\displaystyle 14 \div 8 = 1 \textrm{ R }6\)

The quotient is less than 8, so it will be the first digit. The base-eight equivalent of 117 is \(\displaystyle 165_{\textrm{eight}}\)

The easiest way to explain this is by example. 

We can use 16, since \(\displaystyle 16 \div 9 = 1 \textrm{ R } 7\)

\(\displaystyle 16 ^{2 } = 256\)

\(\displaystyle 256 \div 9 = 28 \textrm{ R } 4\)

This makes 4 the correct choice.

The product of a group of nonzero numbers is negative if and only if an odd number of these factors is negative. This occurs in each of these scenarios except for one - all of the numbers (eight) being negative.

Any integer that is divisible by 5 must end in a 0 or a 5, so any integer which, when divided by 5, yields remainder 3 must end in 3 or 8. Since the cubes of 3 and 8 are 27 and 512, the cube of an integer that ends in one of these digits must end in 2 or 7, meaning that, when divided by 5, the remainder will be 2.

Since the numerator is always 1 greater than the denominator, we know that for large enough values of \(\displaystyle x\), it's never going to be an integer (one will be even, other odd).  In fact, there are only 2 cases where this can be done.  The first is dividing 0.  Since 0 is divisible by every number, if the numerator is 0, then we will still get an integer.  Thus one answer is \(\displaystyle x=1 \Rightarrow \frac{(1)-1}{(1)-2}=\frac{0}{-1} = 0\)

The other answer occurs as a special case as well.  We can divide any number by 1 evenly, so when the denominator is 1 we get an integer:

\(\displaystyle x=3 \Rightarrow \frac{(3)-1}{(3)-2} = \frac{2}{1} = 2\)

In every other case, we will have a non-integer.

If a number has a 5 in its base-six representation, then the number divided by 6 yields a remainder of 5. We test each of these five numbers by dividing each by 6 and noting the remainder:

\(\displaystyle 543 \div 6 = 90 \textrm{ R } 3\)

\(\displaystyle 643 \div 6 = 107 \textrm{ R } 1\)

\(\displaystyle 743 \div 6 = 123 \textrm{ R } 5\)

\(\displaystyle 843 \div 6 = 140 \textrm{ R } 3\)

\(\displaystyle 943 \div 6 = 157 \textrm{ R } 1\)

Among the five choices, only 743 yields remainder 5 when divided by 6, so this is the correct choice.

We can divide this group number into two parts. The first part is \(\displaystyle \left \{ -155, 155\right \}\) and the second part is \(\displaystyle \left \{ 156, 160\right \}\).

The first group is symmetrical, so the sum of this group of numbers is 0. Now for the second part there are only two numbers left according to the question (sum of odd numbers), which is 157 and 159.

Therefore, the answer is \(\displaystyle 157+159=316\).