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Number and Operations Session 6: Solutions
 
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A 
Homework

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Solutions for Session6, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8


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.

<< back to Problem B1


 

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.

<< back to Problem B2


 

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.

<< back to Problem B3


 

Problem B4

a. 

It may, if we can determine what column that number is in. That is, if the number is not one more or one less than a multiple of 6, then it is not prime. To do this, we can check divisibility by 2 and 3. The number is not divisible by 2 (its units digit is not even). To check if it is divisible by 3, add the digits: 9 + 4 + 3 + 7 + 8 + 7 + 5 + 8 + 9 = 60, which is a multiple of 3. So this nine-digit number will be in one of the crossed-out columns and thus is not prime.

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.

<< back to Problem B4


 

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.

<< back to Problem B5


 

Problem B6

a. 

The number 257 is not divisible by 2, 3, 5, 7, 11, or 13. It is prime.

b. 

The greatest prime number we need to check is 13, since it is the largest prime number less than the square root of 257 (which is just over 16).

<< back to 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.

<< back to Problem B7


 

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