Since  divided by 8 yields remainder 5, for some integer quotient ,

This can be rewritten as

Thus,  divided by 4 yields quotient , which is an integer, and remainder 1. The quantities are equal.

Leaving out 60 itself, the factors of 60 are . The sum of all of these factors can easily be seen to exceed 60, since the sum of the three largest factors is

.

This makes (B) greater.

An integer is a multiple of 2 if and only if it ends in 2, 4, 6, 8, or 0; it is a multiple of 5 if and only if it ends in a 0 or 5. We can immediately eliminate these integers, leaving us with this set:

Of these integers, the multiples of 3 are 123, 129, 141, and 147, leaving the set:

The correct response is eight.

An integer is a multiple of 2 if and only if it ends in 2, 4, 6, 8, or 0; it is a multiple of 5 if and only if it ends in a 0 or 5. We can immediately eliminate these integers, leaving us with this set:

We eliminate the multiples of 3, which are 63, 69, 81, 87, 93, and 99:

We then eliminate the multiples of 7, which are 77 and 91:

.

This leaves eight elements.

Add the factors of each number (except for the number itself) and compare to the number:

In each case, the sum of the factors is less than the number, so none of the integers given are abundant.

Add the factors of each number (except for the number itself) and compare to the number:

26, 46, and 86 all have factor sums less than themselves, so the correct response is "three".

An integer is a multiple of 2 if and only if it ends in 2, 4, 6, 8, or 0; it is a multiple of 5 if and only if it ends in a 0 or 5. We can immediately eliminate these integers in each set. 

In the set given in (A), we are left with 

Eliminating the remaining multiples of 3, which are 111, 117, 123, and 129, we are left with

,

a set with eight elements.

Similarly, in the set given in (B), we are left with

.

Eliminating the remaining multiples of 3, which are 201, 207, 213, and 219, we are left with 

,

a set with eight elements.

The quantities are equal.

An integer is a multiple of 2 if and only if it ends in 2, 4, 6, 8, or 0; it is a multiple of 5 if and only if it ends in a 0 or 5. We can immediately eliminate these integers in each set. 

In the set given in (A), we are left with 

Eliminating the remaining multiples of 3, which are 141, 147, 153, and 159, we are left with

Of the remaining numbers, 133 is the only multiple of 7; we remove it, leaving the set

This leaves a set with seven elements.

In the set given in (B), we are left with 

Eliminating the remaining multiples of 3, which are 231, 237, 243, and 249, we are left with

Of the remaining numbers, 259 is the only multiple of 7; we remove it, leaving the set

This leaves a set with seven elements.

The quantities are equal.

Given that , it follows that 25 is a multiple of 125; therefore, the answer choice, "It is a multiple of 25" is the correct answer. 

125 does not have a real-number square root, so it is not the square of a real number (real numbers are those numbers, both rational and irrational, found on a number line). It is also not divisible by 75, and it is smaller that , which is equal to .

First, we must solve for 

While 64 is divisible by 4, 8, and 16, it is not divisible by 7; therefore, 7 is not a factor of 64 and is thus the correct answer.