Each digit of a base six number has a place value that is a power of six. The lowest four powers of 6, beginning with 1, are \(\displaystyle 6^{0} = 1,6^{1} = 6, 6^{2} = 36, 6^{3} = 216\), so

\(\displaystyle 5432_{\textrm{six}} = 5 \times 216 + 4 \times 36 + 3 \times 6 + 2\)

\(\displaystyle = 1,080+ 144 + 18 + 2 = 1,244\)

The square root of an integer is irrational if it is not an integer itself. As can be calculated, all three of these square roots are not integers, so none are rational.

0 and 1 are both integers and are therefore rational numbers. \(\displaystyle \pi\) is known to be an irrational number, since there are no two integers whose quotient is equal to \(\displaystyle \pi\). "Two" is correct.

The whole numbers comprise 0 and the positive integers. \(\displaystyle -17\) is a negative integer, so it belongs to the set of integers, but not the set of whole numbers. Also, all integers are rational by definition, since every integer can be expressed as the quotient of two integers (the integer divided by 1, for example). Therefore, \(\displaystyle -17\) is an integer and a rational number, but not a whole number, and the correct response is two.

We look for ways to write 20 as the product of three integers, then we find the sum of the integers in each situation. They are:

\(\displaystyle 1 \times 1 \times 20\)

Sum: \(\displaystyle 1 + 1 + 20 = 22\)

\(\displaystyle 1 \times 2 \times 10\)

Sum: \(\displaystyle 1+ 2 + 10 = 13\)

\(\displaystyle 1 \times 4 \times 5\)

Sum: \(\displaystyle 1 + 4 + 5 = 10\)

\(\displaystyle 2 \times 2 \times 5\)

Sum: \(\displaystyle 2 + 2 + 5 = 9\)

9, 10, and 13 are among the possible sums, but 15 is not, so that is the correct response.

The whole numbers comprise 0 and the positive integers. \(\displaystyle -17\) is a negative integer, so it belongs to the set of integers, but not the set of whole numbers. Also, all integers are rational by definition, since every integer can be expressed as the quotient of two integers (the integer divided by 1, for example). Therefore, \(\displaystyle -17\) is an integer and a rational number, but not a whole number, and the correct response is two.

20 can be factored as:

I) \(\displaystyle 1 \times 20\)

II) \(\displaystyle 2 \times 10\)

III) \(\displaystyle 4 \times 5\)

The positive difference of the factors can be any of:

\(\displaystyle 20 - 1 = 19\)

\(\displaystyle 10 - 2 = 8\)

\(\displaystyle 5 - 4 = 1\)

Of the four choices, only 8 is possible.

The number of subsets of a set with \(\displaystyle N\) elements is \(\displaystyle 2^{N}\), so this eight-element set has \(\displaystyle 2^{8} = 256\) subsets.

A set with \(\displaystyle N\) elements has \(\displaystyle 2 ^{N}\) subsets. Since \(\displaystyle 64 = 2 ^{6}\), a set with 64 subsets has 6 elements.

We need to find ways to factor 32 such that the three factors are different, and then find the sum of those factors in each case.

32 can be factored as the product of three integers in five differrent ways:

I) \(\displaystyle 1 \times 1 \times 32\) 

II) \(\displaystyle 1 \times 2 \times 16\) 

III) \(\displaystyle 1 \times 4 \times 8\)

IV) \(\displaystyle 2 \times 2 \times 8\)

V) \(\displaystyle 2 \times 4 \times 4\)

Of the five ways, only the second and third involve three distinct factors.

In Case II, the sum of the factors is 

\(\displaystyle 1 + 2 + 16 = 19\).

In Case III, the sum of the factors is 

\(\displaystyle 1 + 4 + 8 = 13\).

The only possible correct response is 19.