Open QA 23

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course Mth 151

7/27 12

Question: `q001. There are twelve questions in this assignment.The number 12 is evenly divisible by 1, 2, 3, 4, 6, and 12. We say that 1, 2, 3, 4, 6 and 12 are the divisors of 12. Each of these divisors can be multiplied by another to get 12. e.g., 2 * 6 = 12, 1 * 12 = 12, 3 * 4 = 12. List the numbers from 2 to 20 and list all the divisors of each.

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Your solution:

The divisors of 2 are 1 and 2.

The divisors of 3 are 1 and 3.

The divisors of 4 are 1, 2 and 4.

The divisors of 5 are 1 and 5.

The divisors of 6 are 1, 2, 3 and 6.

The divisors of 7 are 1 and 7.

The divisors of 8 are 1, 2, 4, hence 8.

The divisors of 9 are 1, 3 and 9.

The divisors of 10 are 1, 2, 5 and 10.

The divisors of 11 are 1 and 11.

The divisors of 12 are 1, 2, 3, 4, 6, 12.

The divisors of 13 are 1 and 13.

The divisors of 14 are 1, 2, 7 and 14.

The divisors of 15 are 1, 3, 5 and 15.

The divisors of 16 are 1, 2, 4, 8 and 16.

The divisors of 17 are 1 and 17.

The divisors of 18 are 1, 2, 3, 6, 9 and 18.

The divisors of 19 are 1 and 19.

The divisors of 20 are 1, 2, 4, 5, 10 and 20

3

Ok

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Question: `q002. Some of the numbers you listed have exactly two divisors. Which are these?

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Your solution:

2,3,5,7,11,13,17,19

3

Ok

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Question: `q003. These numbers with exactly two divisors are called prime numbers. List the prime numbers between 21 and 40.

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Your solution:

counting number except 1 has at least 2 divisors, since every counting number is divisible by itself and by 1. Therefore if a counting number is divisible by any number other than itself and 1 it has more than 2 divisors and is therefore prime. This means that 23,29,31, and 37 would be prime

2

Ok

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Question: `q004. The primes through 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37. Twin primes are consecutive odd numbers which are both prime. Are there any twin primes in the set of primes through 40?

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Your solution:

17 and 19. 5,7,11,13

2

Ok

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Question: `q006. We can prove that 89 is prime as follows:

89 is odd and is hence not divisible by 2.

If we divide 89 by 3 we get a remainder of 2 so 89 is not divisible by 3.

Since 89 is not divisible by 2 it cannot be divisible by 4 (if we could divide it evenly by 4 then, since 2 goes into 4 we could divide evenly by 2; but as we just saw we can't do that).

Since 89 doesn't end in 0 or 5 it isn't divisible by 5.

If we divide 89 by 7 we get a remainder of 5 so 89 is divisible by 7.

Since 89 isn't divisible by 2 it isn't divisible by 8, and since it isn't this will by 3 it isn't divisible by 9.

At this point it might seem like we have a long way to go--lots more numbers to before get to 89. However it's not as bad as it might seem.

For example once we get past 44 we're more than halfway to 89 so the result of any division would be less than 2, so there's no way it could be a whole number. So nothing greater than 44 is a candidate.

We can in fact to even better than that. If we went even as far as dividing by 10, the quotient must be less than 10: since 10 * 10 is greater than 89, it follows that 89 / 10 must be less than 10. So if 10 or, more to the point, anything greater than 10 was going to divide 89 evenly, in the result would be one of the numbers we have already unsuccessfully tried.

It follows that after trying without success to divide 89 by all the numbers through 9, which we didn't really have to try anyway because 9 is divisible by 3 which we already checked, we are sure that 89 has to be prime.

What is the largest number you would have to divide by to see whether 119 is prime?

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Your solution:

7 is the largest number we have to try because once we get to 7 we only have to try 11 which 11*11 is greater than 119 which would not work

3

Ok

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Question: `q007. Precisely what numbers would we have to try in order to determine whether 119 is prime?

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Your solution:

2 3 5 7

2

Ok

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Question: `q008. Is 119 prime?

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Your solution:

Yes it is divisible by more than just one and itself

3

Ok

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Question: `q009. We can 'break down' the number 54 into the product 2 * 27, which can further be broken down to give us 2 * 3 * 9, which can be broken down one more step to give us 2 * 3 * 3 * 3.

We could have broken down 54 into different way as 6 * 9, which could have been broken down into 6 * 3 * 3, which can be broken down one more step to give us 2 * 3 * 3 * 3.

No matter how we break 54 down into factors, the process ends with a single factor 2 and 3 repetitions of the factor 3.

Break down each of the following in this manner, until it is not possible to break it down any further: 63, 36, and 58.

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Your solution:

63 = 9 * 7 = 3 * 3 * 7, 36 = 9 * 4 = 3 * 3 * 4 = 3 * 3 * 2 * 2, which we rearrange in increasing order of factors as 2 * 2 * 3 * 3, and 58 = 2 * 29

3

Ok

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Question: `q010. The results that we obtained in the preceding exercise are called the 'prime factorizations' of the given numbers. We have broken the numbers down until we are left with just prime numbers. Why is it that this process always ends with prime numbers?

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Your solution:

break the numbers down until they can't be broken down any further. If any number in the product is not prime, then it has more than two factors, which means it has to be divisible by something other than itself or 1. In that case we would have to divide it by that number or some other. So the process cannot end until all the factors are prime.

3

Ok

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Question: `q011. Find the prime factorization of 819, then list all the factors of 819.

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Your solution:

In addition to 1 and 819, the factors of 819 will include

3, 7, 13, all of the prime factors

3

Ok

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Question: `q012. List all the factors of 168.

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Your solution:

2,3,7

1

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