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Secondary 1 | Maths
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Need help with question 1,2,3,4,5
Equations and answers required
Date Posted: 6 years ago
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Hi Nicholas. Let me help you with just one question, Q2. The rest are very similar, so after you have understood this, you will be able to do the rest.
Things to note:
Prime Numbers - numbers divisible by itself and 1 ONLY
Perfect Square - all factors have even powers eg 2, 4, 6...etc
Perfect Cube - all factors have powers in multiples of 3 eg. 3, 6, 9...etc
When you are asked to leave a value as a product of its prime factors, the question is asking you to leave you answers in index notation. Break down the value bit by bit, that is to say, divide the value by prime numbers until you cannot divide it anymore, or the remaining value is a prime number itself. When that happens, group all the similar prime numbers together, giving it a power equivalent to the number of that factor present.
For Part (b), the question is telling you that 360k, which is 360 x k, is a multiple of k. This multiple is unique in the sense that all prime numbers have the above-mentioned powers attached.
All you have to do is to compare it to your answer in (a). 360k (which is 360 x k), has the value of 360 in it. As such, ALL PRIME FACTORS listed in Part (a) will be present in 360k. Take for example, the prime number 5. There is only ONE value of 5 in 360. But as 360k is a perfect square, the closest value in 360k would be 5 x 5 (which is 5-squared). The extra 5 would be found in k.
When we apply this same logic to a perfect cube, the closest value in 360k for the prime number 5, would be 5 x 5 x 5 (5-cube). As such, the 2 extra 5's are found in k.
By comparing your answer in Part (a), fill in all prime numbers until they reach your goal of having a squared or cubed value. The values you add to achieve that, would then be found in k. Multiply them all together, and you will find the value of k.
Things to note:
Prime Numbers - numbers divisible by itself and 1 ONLY
Perfect Square - all factors have even powers eg 2, 4, 6...etc
Perfect Cube - all factors have powers in multiples of 3 eg. 3, 6, 9...etc
When you are asked to leave a value as a product of its prime factors, the question is asking you to leave you answers in index notation. Break down the value bit by bit, that is to say, divide the value by prime numbers until you cannot divide it anymore, or the remaining value is a prime number itself. When that happens, group all the similar prime numbers together, giving it a power equivalent to the number of that factor present.
For Part (b), the question is telling you that 360k, which is 360 x k, is a multiple of k. This multiple is unique in the sense that all prime numbers have the above-mentioned powers attached.
All you have to do is to compare it to your answer in (a). 360k (which is 360 x k), has the value of 360 in it. As such, ALL PRIME FACTORS listed in Part (a) will be present in 360k. Take for example, the prime number 5. There is only ONE value of 5 in 360. But as 360k is a perfect square, the closest value in 360k would be 5 x 5 (which is 5-squared). The extra 5 would be found in k.
When we apply this same logic to a perfect cube, the closest value in 360k for the prime number 5, would be 5 x 5 x 5 (5-cube). As such, the 2 extra 5's are found in k.
By comparing your answer in Part (a), fill in all prime numbers until they reach your goal of having a squared or cubed value. The values you add to achieve that, would then be found in k. Multiply them all together, and you will find the value of k.
Date Posted:
6 years ago
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