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**In This Part: Locating Prime Numbers**

In this part, we’ll continue a mathematics tradition begun by Eratosthenes of Cyrene (276-194 B.C.E.) — the same person who’s known for accurately estimating the diameter of the Earth based on the shadow cast by the Sun’s light.

Eratosthenes worked out a method, now called the “Sieve of Eratosthenes,” to collect all the prime numbers and allow all composites (multiples of prime numbers) to “drain through.” He used a grid that looked like what we now call the 100 board — the first row is 1-10, the second row 11-20, etc. This grid does locate the prime numbers, but it does not help us understand where to look for them. If you try looking for prime numbers in this grid, you will discover that it’s not so easy to locate them in a systematic way:

In the following activity, we will use a different grid to locate the prime numbers. This grid has only six columns, starting with the numbers 2 through 7. As you will see, such positioning of numbers will make the patterns more noticeable and consequently will be more helpful in answering the question of where the prime numbers are located.

Copy the grid above or print a PDF version.

**Problem B1
**Circle the 2, which is a prime number. Next, cross out all the multiples of 2, as they are not prime numbers.

**Problem B2
**Next, circle the smallest remaining prime number (i.e., 3). Cross out all the multiples of 3, as they are not prime.

**a. **Again imagine that the grid goes on forever. Present a convincing argument for the fact that all the numbers in the second and fifth columns are multiples of 3 and would thus be crossed out. (Of course, the fifth column is already gone because it contains multiples of 2!)

**b. **Could multiples of 3 be located in any other column?

**Problem B3
**Again, circle the smallest remaining prime number (i.e., 5) and cross out all the multiples of 5, as they are not prime. Notice that these multiples are not all located in particular columns, so crossing them out is not as easy as before. Continue in a similar manner until all the numbers on the grid are either circled or crossed out.

Examine the grid and imagine that it extends to infinity. Where on the extended grid should you look to find prime numbers greater than 3?

**Video Segment
**Watch how Professor Findell and the participants worked on Problems B1-B3 in this video segment.

Are you convinced that the conjecture about locating prime numbers is true? Try extending the table by a few more rows to see if the rule still applies.

You can find this segment on the session video approximately 10 minutes and 56 seconds after the Annenberg Media logo.

**In This Part: Necessary and Sufficient Conditions
**The previous activities illustrate the difference between a necessary and a sufficient condition. We have shown that every prime greater than 3 is located in either the fourth or sixth columns of our grid. This means that it is

However, we also found numbers in those columns that are composites (not prime); thus, that location is not a *sufficient* condition for a number to be prime. This type of thinking is very useful when analyzing relationships in mathematics.

To give another example: For a number to be divisible by 6, it is necessary, but not sufficient, that it is divisible by 3. Conversely, for a number to be divisible by 3, it is sufficient, but not necessary, that it is divisible by 6.

It is *necessary* for the rest of the primes to fall in either the fourth or sixth columns, but that is not a *sufficient* condition.

**Problem B4
**

**In This Part: Is This Number Prime?
**Finding factors and checking if a large number is prime remains one of the most time-consuming tasks in mathematics. Even very powerful computers cannot do this task quickly. For this reason, prime numbers are very useful in cryptography. Secret messages are sometimes coded using large numbers that are the product of two large prime numbers.

Here are the rules you can use to find out whether a particular number is prime:

**• **Pick a number n.

**• **Start with the least prime number, 2. See if 2 is a factor of your number. If it is, your number is not prime.

**• **If 2 is not a factor, check to see if the next prime, 3, is a factor. If it is, your number is not prime.

**• **Keep trying the next prime number until you reach one that is a factor (in which case n is not prime), or you reach a prime number that is equal to or greater than .

**• **If you have not found a factor less than or equal to , you can be sure that your number is prime.

Let’s try a number; for example, 97. To check if 97 is a prime number, we start a list of factors, as above:

1 • 97

2 • (no number works here)

3 • (no number works here)

(4 is not prime; it doesn’t need to be checked, because we know 2 didn’t work)

5 • (no number works here)

(6 is not prime; it doesn’t need to be checked, because 2 and 3 didn’t work)

7 • (no number works here)

(8 is not prime; it doesn’t need to be checked, because 2 and 4 didn’t work)

(9 is not prime; it doesn’t need to be checked, because 3 didn’t work)

This brings us to 10. Ten is greater than the square root of 97, and therefore its partner on the right would have to be less than 10. Since we’ve already checked every number less than 10, we know that none of them are factors of 97. Therefore, 97 is prime.

**Problem B5**

**a. **Why don’t you need to check any prime numbers greater than 11 to see if 127 is prime?

**b. **Can you explain the rule for where to “stop?”

**Problem B6
**

**Problem B7
**Use this method to determine if 359 is prime.

**Video Segment
**In this video segment, Liz and Jeanne try to determine whether a large number is prime. They apply divisibility rules and computational shortcuts to solve the problem. Watch this segment after you’ve solved Problems B5-B7.

Were the methods you came up with similar or different?

You can find this segment on the session video approximately 16 minutes and 50 seconds after the Annenberg Media logo.

In Part A of this session, we looked for prime factors of numbers. Using this method, the prime factorization of 72 is 2 • 2 • 2 • 3 • 3, or 2^{3} • 3^{2}. In Part B, we listed all the factors of the number 72 in ascending order: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.

Notice that these two lists are very different. The prime factorization, which the fundamental theorem of arithmetic says is unique, lists only the prime factors of 72 and lists each factor as many times as it appears.

In contrast, the list of all factors lists all numbers that are factors of 72, many of which are not prime. When doing problems, make sure to think about this distinction.

**Problem B1
a. **All the numbers in the first, third, and fifth columns end in 2, 4, 6, 8, or 0, so they are all multiples of 2. (Refer to Session 5 if you need a refresher.)

Another convincing argument is that each number in a column is six more than the number directly above it. In other words, to move down a column, you add 6. Since 2 divides 6, 2 will divide any even number plus 6. So, as you move down a column, you will continue to get multiples of 2.

Yet another convincing argument is that because there are six numbers in each row, and 2 is a factor of 6, then each row has the multiples of 2 in the same position.

**b. **No. The number after an even number (or multiple of 2) is always odd, so any number in a column directly after an even number cannot also be even. As in the previous answer, all numbers in the second, fourth, and sixth columns are multiples of 6 more than the top number in the column. So the numbers in the second, fourth, and sixth columns are all a multiple of 6 more than an odd number and are therefore odd numbers themselves.

**Problem B2
**

**a. **All the numbers in the second and fifth columns have digits that add to a multiple of 3. Thus, referring back to divisibility tests, we know that all those numbers are divisible by 3.

Another argument is that because 3 is a factor of 6, each row has the multiples of 3 in the same position. Numbers in the second column are all a multiple of 6 more than 3 and are thus multiples of 3. Numbers in the fifth column are all multiples of 6.

**b. **No. In six consecutive numbers, there cannot be more than two multiples of 3. Numbers that are a multiple of 6 more than 2, 4, 5, or 7 will not be multiples of 3. In other words, if you take a column where every number is a multiple of 3, then every number in the column before that is one less than a multiple of 3 — thus, it cannot be a multiple of 3. Likewise, every number in the column after it is one more than a multiple of 3 and thus cannot be a multiple of 3.

**Problem B3
**The only numbers that can be prime numbers greater than 3 are the numbers that have not been crossed out yet. All of the remaining prime numbers must be located in the fourth or sixth columns, because the other columns are all crossed out.

In particular, the first number we have not crossed out yet must be prime. After crossing out multiples of 2 and 3, the first number we have not crossed out yet is 5.

The only two columns that have numbers not already crossed out are on either side of the column that contains all multiples of 6. That means that any prime number greater than 3 has to be either one more or one less than a multiple of 6.

**Problem B4
**

**b. **The number 532,391 is not divisible by 2 or 3, and therefore is not a multiple of 6. To further check its location, you need to check the divisibility by 2 of the numbers one more and one less than 532,391; both clearly are. Then you need to check divisibility by 3 of the numbers one more and one less. A quick check shows that the digits of 532,392 sum to 24, so it’s divisible by 3. This means that the number 532,392 is divisible by 6 and thus is located in the fifth column. The number 532,391 is located in the fourth column, and it may or may not be a prime.

**Problem B5
a. **We only need to check prime numbers up to the square root of the given number. The square root of 127 is between 11 (the square root of 121) and 12 (144), so we only need to check the prime numbers up to 11.

**b. **By definition of square roots *n* will factor as • . Now, think of finding other factors from this “middle point.” If you change the first number to be something larger than , the second factor must get smaller to make the product stay constant at n. So you’re guaranteed that when you factor n into a product of exactly two numbers, at least one of the two will be less than or equal to . Since this is true of any pair of two factors, it’s certainly true if we restrict one of the factors to be a prime as well.

**Problem B6
**

**Problem B7
**We must check all prime numbers up to 17 (since the square root of 359 is just under 19). Since 359 is not divisible by 2, 3, 5, 7, 11, 13, or 17, it is prime.