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primary 6 | Maths
| Data Analysis
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Date Posted: 6 months ago
Views: 67
There is a minimum of 5 factors, simply because B could contain A as a factor.
Take for example, A = 3 and B = 27.
Check that B contains only 1, 3, 9 and 27 as factors.
The product of this is 81, and 81 contains 1, 3, 9, 27 and 81.
(Using higher level ideas)
The minimum of 5 factors can only be achieved when B = n and A takes the form n^3 where n is a prime number.
A x B = n^4, and there are only five factors 1, n, n^2, n^3 and n^4.
Take for example, A = 3 and B = 27.
Check that B contains only 1, 3, 9 and 27 as factors.
The product of this is 81, and 81 contains 1, 3, 9, 27 and 81.
(Using higher level ideas)
The minimum of 5 factors can only be achieved when B = n and A takes the form n^3 where n is a prime number.
A x B = n^4, and there are only five factors 1, n, n^2, n^3 and n^4.
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Keep in mind that pm means prime number and a prime number is a number that has only 2 factors, itself and 1.
Date Posted:
6 months ago
There is a flaw with this working.
Try A = 2 and B = 8.
Try A = 2 and B = 8.
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