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Chapter 5: Problem 43

Find the prime factorization of each composite number. 85,800

Short Answer

Expert verified
The prime factorization of 85,800 is \(2^3 \times 3 \times 5^2 \times 11^1 \times 13^1\).

Step by step solution

01

Understand the concept of Prime Factorization

Prime factorization is finding which prime numbers multiply together to make the original number. Prime numbers are: numbers that have only two factors - one and the number itself.
02

Factorize the number

Let's start by finding the prime factors of 85,800. Start by dividing the number by the smallest prime number, which is 2. 85,800 梅 2 = 42,900. Repeat the process with the result. 42,900 梅 2 = 21,450. Repeat one more time: 21,450 梅 2 = 10,725. Now, the number 10,725 is not divisible by 2, try the next prime number which is 3. It is not divisible by 3 as well. The next prime number, 5, can divide 10,725 to give 2,145. Continue with the prime numbers and 2,145 梅 5 = 429. Keep going, and next, 429 梅 11 = 39 and then 39 梅 3 = 13. At this point, we are left with 13 which is a prime number itself.
03

Write down the Prime Factorization result

So, the prime factorization of 85,800 is: \(2^3 \times 3 \times 5^2 \times 11^1 \times 13^1\). We see the numbers 2, 3, 5, 11, and 13 appearing in the factorization because these are the prime numbers that divide into 85,800.

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Most popular questions from this chapter

Company A pays $$\$ 24,000$$ yearly with raises of $$\$ 1600$$ per year. Company B pays $$\$ 28,000$$ yearly with raises of $$\$ 1000$$ per year. Which company will pay more in year 10 ? How much more?

Use a calculator with a square root key to find a decimal approximation for each square root. Round the number displayed to the nearest \(\mathbf{a}\). tenth, b. hundredth, \(\mathbf{c}\) thousandth. \(\sqrt{173}\)

There are two species of insects, Magicicada septendecim and Magicicada tredecim, that live in the same environment. They have a life cycle of exactly 17 and 13 years, respectively. For all but their last year, they remain in the ground feeding on the sap of tree roots. Then, in their last year, they emerge en masse from the ground as fully formed cricketlike insects, taking over the forest in a single night. They chirp loudly, mate, eat, lay eggs, then die six weeks later. (Source: Marcus du Sautoy, The Music of the Primes, HarperCollins, 2003) a. Suppose that the two species have life cycles that are not prime, say 18 and 12 years, respectively. List the set of multiples of 18 that are less than or equal to 216 . List the set of multiples of 12 that are less than or equal to 216. Over a 216-year period, how many times will the two species emerge in the same year and compete to share the forest? b. Recall that both species have evolved prime-number life cycles, 17 and 13 years, respectively. Find the least common multiple of 17 and 13 . How often will the two species have to share the forest? c. Compare your answers to parts (a) and (b). What explanation can you offer for each species having a prime number of years as the length of its life cycle?

Shown in the figure is an 8-hour clock and the table for clock addition in the 8-hour clock system. $$ \begin{array}{|l|l|l|l|l|l|l|l|l|} \hline \oplus & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} \\ \hline \mathbf{0} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathbf{1} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 0 \\ \hline \boldsymbol{2} & 2 & 3 & 4 & 5 & 6 & 7 & 0 & 1 \\ \hline \mathbf{3} & 3 & 4 & 5 & 6 & 7 & 0 & 1 & 2 \\ \hline \mathbf{4} & 4 & 5 & 6 & 7 & 0 & 1 & 2 & 3 \\ \hline \mathbf{5} & 5 & 6 & 7 & 0 & 1 & 2 & 3 & 4 \\ \hline \mathbf{6} & 6 & 7 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathbf{7} & 7 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \end{array} $$ a. How can you tell that the set \(\\{0,1,2,3,4,5,6,7\\}\) is closed under the operation of clock addition? b. Verify the associative property: $$ (4 \oplus 6) \oplus 7=4 \oplus(6 \oplus 7) \text {. } $$ c. What is the identity element in the 8-hour clock system? d. Find the inverse of each element in the 8-hour clock system. e. Verify two cases of the commutative property: \(5 \oplus 6=6 \oplus 5\) and \(4 \oplus 7=7 \oplus 4\).

Insert one pair of parentheses to make each calculation correct. \(8-2 \cdot 3-4=10\)
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