The common difference for this arithmetic sequence is . To find the next term in the sequence, subtract that value from the last given value.

The sequence is subtracting  for each term. The next term can be found by subtracting  from :

The common difference for this arithmetic sequence is . Take away  from  to get the next term.

The common difference for this sequence is . Subtract  from the last given term to find the next term.

 is the complement of , the set of all elements in the universal set not in . In the diagram, it is represented by all elements not inside the circle that represents . This is the set:

 is the union of the sets  and  - that is, the set of all elements in either  or .  is the complement of set  - that is, the set of elements in  not in ;  is defined similarly. Therefore, we need all of the elements either outside of  or outside of . The excluded elements will be those inside both sets, which, by the diagram, can be seen to be 3, 5, 13, and 14. Therefore,

 .

The region in which 2,431 appears depends on the sets of which 2,431 is an element, which in turn depends on which of 7, 11, and 13 divides it evenly:

2,431 is a multiple of 11 and 13, but not 7, so 2,431 is in  and , but not . We look for a number among the choices that is in  and , but not  - that is, a number divisible by 11 and 13 but not 7.

, so 2,772 is divisible by 7. We can eliminate it.

, so 2,184 is divisible by 7. We can eliminate it.

, so 3,081 is not divisible by 11, and we can eliminate it.

, so 2,409 is not divisible by 13, and we can eliminate it.

However:

2,145 is divisible by 11 and 13 but not 7, so this is the correct choice.

Cathy is in the Honor Society, meaning that she is in set ; she turned 18 after election day, so she is not in set ; she is taking a French course, so she is in set . She is in set , represented by region 2.

The shaded region is , which will comprise all of the numbers that are multiples of 10 and of either or both 8 or 9.

 will comprise multiples of 10 and 8 - that is, multiples of . Since , 

 will comprise multiples of 10 and 9 - that is, multiples of . Since , 

To find , we first find .

, the set of numbers that are multiples of 8, 9, and 10. , so we look for multiples of 360, of which there are two under 1,000 (360 and 720). 

The shaded region is inside  and , and outside of , meaning that the shaded set represents

Examine  first. Each number in  must be a multiple of 7 and 8; since the two are relatively prime, each number is a multiple of 56.

so seventeen elements are in .

We eliminate all elements in  - that is, all elements that also multiples of 6. These elements are 168, 336, 504, 672, 840 - a total of five. 

This leaves twelve elements in .