Source: NCTM Mathematics Teacher, August 2006

**Solution**

We assume that the three numbers are not consecutive but in some random order in the geometric progression with first term = and common ratio = . Given that their product =, the number of factors of . They are . The sum of the three numbers is small, so we concentrate on the smaller factors like . By trial and error we find that the three numbers are .

Verification

We either have a decreasing geometric progression with common ratio = or an increasing geometric progression with common ratio . The largest of the three numbers is .

**Answer**:

**Alternative solution
**If we assume that the three numbers are consecutive, they can be written as where is an integer and is the common ratio.

Solving for using the quadratic formula yields or . Hence, the three numbers are or .

Source: NCTM Mathematics Teacher, August 2006

**Solution**

If leaves a remainder of , the possible values of are . The integers form an arithmetic progression with first term = and common difference = . The value of the th term of is

Similarly, they form arithmetic progressions with first terms = and common differences = . The values of their respective th terms are given below

Note that the are multiples of . The smallest such multiple is the LCM of which is . Thus, the smallest positive integer that when divided by leaves a remainder of respectively is .

**Answer**:

Source: NCTM Mathematics Teacher, August 2006

**Solution**

Consider , the set of four-digit positive integers divisible by and , the set of four-digit positive integers divisible by . The count of and the count of .

Some integers like and appear in both sets so we subtract them as duplicates from the total count. Since , starting from the duplicates occur every th integer in set or every rd integer in set . If we use set , the count of duplicates = . The number of four-digit positive integers divisible by and/or is .

**Answer**:

**Alternative solution**

If we divide by , we get which means that there are integers that are divisible by . But, if we divide by , we get which is not an integer. So we say there are integers that are divisible by . To avoid this problem we use the floor function = the largest integer less than or equal to for some real number . In our example, .

Number of four-digit integers divisible by

Number of four-digit integers divisible by

Number of four-digit integers divisible by

There are four-digit integers divisible by and/or .

Source: NCTM Mathematics Teacher, August 2006

**Solution**

Draw perpendicular to the bases of the trapezoid through point . Since is parallel to , triangles and are similar. The ratio of the lengths of their corresponding sides equals the square root of the ratio of their areas.

For ease in computation, let for some integer .

Area of trapezoid

Area of triangle

Area of trapezoid

**Answer**: square units

Source: NCTM Mathematics Teacher, August 2006

**Solution**

a) All three coins are the same

Number of ways =

b) Two of the coins are the same

Number of ways =

c) All three coins are different

Number of ways =

Total number of ways =

If we check the values of these ways, the values are all different. There are possible values when three coins are selected from among pennies, nickels, dimes, and quarters.

**Answer**:

Source: NCTM Mathematics Teacher, August 2006

**Solution**

We arrange all the non-negative integers into buckets depending on the value of the remainder when they are divided by . It works the same for negative integers too but we ignore them for this problem.

For example, falls in the bucket because . It is not practical to do the same direct calculation for large value like so we enlist the power of modulo arithmetic.

The remainder when is divided by is .

**Answer**:

Source: NCTM Mathematics Teacher, August 2006

**Solution**

The next prime greater than is . If , then will have the same set of prime divisors as .

**Answer**

Source: NCTM Mathematics Teacher, August 2006

**Solution**

For to be an integer, must be a multiple of .

for some integer

Thus, divides and the quotient is odd. Since does not divide , it must divide and must be odd. The largest -digit that meets the requirement is .

**Answer**:

**Alternative solution**

is a point on the graph of . We want the largest -digit integer such that is a solution. Let for some integer .

For to be an integer, must divide . Since and are relatively prime, must divide . Hence, for some integer .

Source: NCTM Mathematics Teacher, August 2006

**Solution**

Area

Vertical shared edges

Horizontal shared edges

Total shared edges

Area

Vertical shared edges

Horizontal shared edges

Total shared edges

Area

Vertical shared edges

Horizontal shared edges

Total shared edges

From the pattern we derive the number of shared edges in a area

Vertical shared edges

Horizontal shared edges

Total shared edges

Instead of thinking of how many ways can we lay the two defective tiles so that they share an edge, equivalently we can think of how many shared edges there are — each shared edge corresponds to a way of laying the two defective tiles.

Number of desirable outcomes =

Number of possible outcomes = — ways to choose tiles out of

Probability (two defective tiles share an edge) =

**Answer**:

**Alternative Solution 1
**If the first defective tile is placed in a location in an upper row (there are seven upper rows), the second defective tile can occupy the adjacent location directly below it.

Probability =

If the first defective tile is placed in a location in a left column (there are seven left columns), the second defective tile can occupy the adjacent location directly to its right.

Probability =

Probability (two defective tiles share an edge)

**Alternative solution 2
**

The first defective tile may be placed within the interior — probability =

Or in the four edges — probability =

Or in the four corners — probability =

Once the first defective tile is placed, the second defective tile must be placed in an adjacent location for them to share an edge.

If the first defective tile is placed in an interior location, the second defective tile can occupy four possible adjacent locations — probability =

If the first defective tile is placed in an edge location, the second defective tile can occupy three possible adjacent locations — probability = 3/63

If the first defective tile is placed in a corner location, the second defective tile can occupy two possible adjacent locations — probability =

Probability (two defective tiles shared an edge)

Source: NCTM Mathematics Teacher, August 2006

**Solution
**Let denote and denote .

= probability of choosing

= probability of choosing

(a)

(b)

(c)

(d)

(e)

The sequences in increasing likelihood are (c), (e), (a), (b), (d)

**Answer**: (c), (e), (a), (b), (d)

**Alternative solution**

(a)

Number of desirable outcomes =

Number of possible outcomes =

(b)

Number of desirable outcomes =

Number of possible outcomes =

(c)

Number of desirable outcomes =

Number of possible outcomes =

(d)

Number of desirable outcomes =

Number of possible outcomes =

(e)

Number of desirable outcomes =

Number of possible outcomes =

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