Beginning with the third term, each term is the sum of the previous two:

\(\displaystyle 1+3 = 4\)

\(\displaystyle 3+4 = 7\)

\(\displaystyle 4+ 7 = 11\)

\(\displaystyle 7+11 = 18\)

\(\displaystyle 11+18 = 29\)

\(\displaystyle 18 + 29 = 47\), the number in the circle;

\(\displaystyle 2 9+ 47 = 76\), the number in the square.

The number that is added to each term to obtain the next one increases by 2 each time:

\(\displaystyle 4+ 4 = 8\)

\(\displaystyle 8+6 = 14\)

\(\displaystyle 14 + 8 = 22\)

\(\displaystyle 22+10 = 32\)

\(\displaystyle 32 + 12 = 44\)

If we let \(\displaystyle X\) be the number in the circle, then to maintain the pattern,

\(\displaystyle X + 2 = 4\)

so

\(\displaystyle X = 2\). This is the correct choice. 

Each term, beginning with the third, is the sum of the previous two:

\(\displaystyle 4+ 11 = 15\)

\(\displaystyle 11+ 15 = 26\)

\(\displaystyle 15+26 = 41\)

\(\displaystyle 26 + 41 = 67\)

\(\displaystyle 41+ 67 = 108\)

If \(\displaystyle X\) is the term in the circle, then 

\(\displaystyle X+ 4 = 11\)

and

\(\displaystyle X = 7\).

In an arithmetic sequence, each term is obtained by adding the same number to the previous term - the common difference. Since \(\displaystyle 7 - 1 = 6\), 6 is the common difference. The next two terms are:

\(\displaystyle 7+6 = 13\)

\(\displaystyle 13+6 = 19\)

19 replaces the square and is the correct choice.

In a geometric sequence, each term is obtained by muliplying the previous term by the same number - the common ratio. 

Since \(\displaystyle 7 \div 1 = 7\), 7 is the common ratio. The next two numbers in the sequence are 

\(\displaystyle 7 \times 7 = 49\)

\(\displaystyle 49 \times 7 = 343\)

343 replaces the square; of the four choices, 350 comes closest and is the correct response.

There are three ways to choose a meat topping - four minus the one Mike won't eat.

There are five vegetable toppings - six minus the one to which John is allergic - from which to choose three. This is

\(\displaystyle C(5,3)\)

\(\displaystyle = \frac{5!}{(5-3)! 3!}\)

\(\displaystyle = \frac{5!}{2! 3!}\)

\(\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4\cdot 5 }{1 \cdot 2 \cdot 1 \cdot 2\cdot 3 }\)

\(\displaystyle = \frac{ 4 \cdot 5 }{ 1 \cdot 2 }\)

\(\displaystyle = \frac{20}{2}\)

\(\displaystyle = 10\)

The number of ways to make a Special Pizza that leaves out anchovies and mushrooms is 

\(\displaystyle 3 \cdot C(5,3) = 3 \cdot 10 = 30\)

Roger paid $915 rent for each of eight complete months - January through August - for a total of

\(\displaystyle \$ 915 \times 8 = \$7,320\)

He paid the rent for May two days late, so the penalty was

\(\displaystyle \$30 \times 2 = \$60\)

To break the lease, he paid two months' rent, or

\(\displaystyle \$ 915 \times 2 = \$ 1,830\)

Add these, add the prorated September rent, and subtract the returned deposit:

\(\displaystyle \$(7,320+1,830+60+600-400) = \$9,410\)

A number is divisible by 4 if and only if its last two digits form a number which itself is divisible by 4. The numbers 08, 28, 48, 68, and 88 are each divisible by 4, as seen here:

\(\displaystyle 8 \div 4 = 2\)

\(\displaystyle 28 \div 4 = 7\)

\(\displaystyle 48 \div 4 = 1 2\)

\(\displaystyle 68 \div 4 = 17\)

\(\displaystyle 88 \div 4 = 22\)

However, 18, 38, 58, 78, and 98 are not divisible by 4, as seen here:

\(\displaystyle 18 \div 4 = 4 \textup{ R }2\)

\(\displaystyle 38 \div 4 = 9 \textup{ R }2\)

\(\displaystyle 58 \div 4 = 1 4 \textup{ R }2\)

\(\displaystyle 78 \div 4 = 19 \textup{ R }2\)

\(\displaystyle 98 \div 4 = 2 4 \textup{ R }2\)

Therefore, the ways to fill in the box to form a number divisible by 4 are 2, 4, 6, 8, and 0 - five ways.

An integer is divisible by 3 if and only if its digit sum is also divisible by 3. If we let \(\displaystyle N\) be the missing digit, then the digit sum is 

\(\displaystyle 3+9+3+3+ 8 + N + 8 = N + 34\)

We can test each digit from 0 to 9 to see which sums are divisible by 3. We can see that this is the case if:

\(\displaystyle N = 2 : N + 34 = 2 + 34 = 36\)

\(\displaystyle N =5: N + 34 =5+ 34 = 39\)

\(\displaystyle N = 8 : N + 34 = 8 + 34 = 42\)

The other seven digits can be seen to form a digit sum that is not a mulitiple of 3 using the same method.

Therefore, there are three ways to fill in the box to form a multiple of 3.

For each of the four couples, there are two ways to select the table at which to seat each person - the man can be at the left table and the woman can be at the right, or vice versa. This makes \(\displaystyle 2 ^{4} = 16\) ways to make this decision total between the four couples.

Once these four choices are made, for each table, there are 

\(\displaystyle P (4,4) = 4 ! = 1 \times 2 \times 3 \times 4 = 24\)

ways to seat the four chosen persons. 

By the multiplication principle. there will be 

\(\displaystyle 16 \times 24 \times 24 = 9,216\) different arrangements that seat each husband and wife at separate tables.