Out of the ten integers from 41 to 50, the prime numbers are 41, 43, and 47 - therefore, there are three prime numbers and seven composite numbers. Since

,

(B) is greater.

The prime numbers between 50 and 60 are 53 and 59; their sum is . The odd composite numbers between 50 and 60 are 51, 55, and 57; their sum is . Therefore, (B) is greater.

The least prime integer between 60 and 70 is 61, so this is the minimum value of . The least prime integer between 80 and 90 is 83, so this is the minimum value of . Since 

 and ,

then , by the addition property of inequality, 

.

There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of . 97 is the only prime number between 90 and 100, so . 

If , then .

If , then .

If , then .

The  prime numbers between 10 and 100 that end in a 7 are 17, 37, 47, 67, and 97 (27, 57, and 87 all have 3 as a factor; 77 has 7 as a factor).

Their sum is .

There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of . There are two prime numbers between 30 and 40 - 31 and 37 - so these are the possible values of . 

Therefore, we find  for six scenarios:

The possible values of  are given by the set 

.

The greatest prime integer between 30 and 40 is 37, so this is the maximum value of . The greatest prime integer between 20 and 30 is 29, so this is the maximum value of . Since 

 and , 

then, by the addition property of inequality, 

.

The numbers between 100 and 200 that feature 7 as their middle digit are 

We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):

171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves 

By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.

A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is

This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves

Since , we can remove 187; there are no multiples of 13, and we need go no further. The primes are 

.

The set in (B) has six elements and is the set of greater cardinality. (B) is greater.

The numbers between 100 and 200 that feature 3 as their last digit are 

.

There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:

Of the remaining numbers:

No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set

,

a set with five elements.

A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are 

There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:

Of the remaining numbers,

No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set

which also has five elements.

The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.

We show that it cannot be determined whether  is greater than, less than, or equal to 10 by choosing two primes within the given ranges and subtracting.

Case 1:

Case 2:

In each case, , with  and  prime.