Solve problems by finding the prime factors of numbers.
Number Framework Stage 8
On a calculator, 1 ÷ 9 = 0.1111111111 ... yet 456 ÷ 62 500 = 0.007296. It is not easy to see which divisions have recurring decimal answers and which are terminating.
Using Number Properties
Problem: “Write down the prime factorisations of all of these numbers: 8, 20, 100, 25, 16, 125, 500, 128 ...
What do you notice?” (Answer: The only prime factors are 2 or 5.)
Problem: “Using the numbers above as the divisors, find these answers on a calculator:
231 ÷ 20, 29 ÷ 1.6, 89 ÷ 0.25, 14 ÷ 12.5, 73 ÷ 3.2, 45 ÷ 8
What do you notice about all the answers?” (Answer: They all terminate.)
Problem: “Hiria works out 455 ÷ 3 and gets 151.6666666. Discuss what the actual answer is.”
(Answer: It is 151.666666666 ... where 6 repeats forever.)
Write 151.666666666 ... as 151.6 on the board.
Problem: “Find these answers on a calculator and write the answers as recurring decimals:
21.4 ÷ 3, 323.1 ÷ 12, 10.069 ÷ 9, 24 567.9 ÷ 11, 1.023 ÷ 15, 12.045 ÷ 0.06, 12.333 ÷ 0.7
Problem: “How, by looking at the divisor only, can we predict which divisions will have a recurring decimal?”
(Answer: Treating the divisor as a whole number, the only prime factors of the divisor must be 2 or 5 for the decimal to terminate.)
Examples: Predict without a calculator which of these fractions will be recurring in their decimal form: 23/32, 64/65, 21/250, 890/12201
Understanding Number Properties:
The denominator of 24/96 appears to show it is a recurring decimal, but it is not. Why? (Answer: 24/96 = 1/4, which terminates.)
Extension Activity
Problem: “Find 34 ÷ 7 and explain why the answer is a recurring decimal whose repeating cycle cannot be more than 6 digits.” (Answer: 34 ÷ 7 = 4.857142857... This indicates 34 ÷ 7 = 4.857142. The cycle is 6 digits long. It cannot be more than 6 digits because a number ÷ 7 has 7 possible remainders 0, 1, 2, 3, 4, 5, 6 at any stage of the long division. If the remainder is ever 0, the division ceases, so the answer in such a case is not a recurring decimal. So, if the decimal is recurring there are only 6 possible remainders, which must be reused at most after 6 applications of the division process.)