Rubin Posted April 26, 2006 Share Posted April 26, 2006 The factors for 156 are 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156 grouped by last digit {1},{2, 12, 52},{3,13},{4},{6, 26, 156},{78},{39} Fenders are defined as factor enders. For example the fenders (factor enders) of 156 are 1, 2, 3, 4, 6, 8, 9 and that 156 is a 7-fender (that is, it has seven fenders). (Question 1) Show that a number which has 0 and 9 as fenders has at least four more fenders. (Question 2) The smallest 10-fender is less than 700. Find it and explain why it is the smallest. Link to comment Share on other sites More sharing options...

nicobudini Posted April 26, 2006 Share Posted April 26, 2006 Ok, let's see if this helps. (Question 1)Show that a number which has 0 and 9 as fenders has at least four more fenders. If a number has 0 and 9 as fenders then it must be divisible by 10 and 9, because you can't divide by 0 and 9 is the first number that ends with 9. Now: - divisible by 10[math]\Longrightarrow[/math]divisible by 2 and by 5 - divisible by 9[math]\Longrightarrow[/math]divisible by 3 Until now you have 3 more fenders {2,3,5}, and the other is 1. So the fenders are {1,2,3,5}, four at least. Question 2, coming soon Link to comment Share on other sites More sharing options...

.L.I.V. Posted May 29, 2006 Share Posted May 29, 2006 Hi, I have the same maths question and I am in year 7. I have the answer to question 2. The answer is 540. I chose 1,4,5,27 and multiplied all of them together. The 1 for the 1 fender The 4 for the 2 and 4 fender (even numbers are divisible by 2) The 5 for the 5 and 0 fender (even numbers multiplied by 5 results in a multiple of 10) The 27 for the 3, 6, 7, 8 and 9 fender 27 multiplied by an even number would result it being divisible by 6 and 18. The factors of 6 is 3 and 2, the factors of 18 is 3, 3, 2. I hope it is helpful. If you could, could you help me with a question? (Question 3) Find three 9-fenders less than 1000 with different sets of fenders. I already have 360, and I could have had 720 and 900 if the question had not asked for the fender sets to be different... Link to comment Share on other sites More sharing options...

yr7student Posted June 10, 2006 Share Posted June 10, 2006 Ok' date=' let's see if this helps. If a number has 0 and 9 as fenders then it must be divisible by 10 and 9, because you can't divide by 0 and 9 is the first number that ends with 9. Now: - divisible by 10[math']\Longrightarrow[/math]divisible by 2 and by 5 - divisible by 9[math]\Longrightarrow[/math]divisible by 3 Until now you have 3 more fenders {2,3,5}, and the other is 1. So the fenders are {1,2,3,5}, four at least. Question 2, coming soon Your explanation is wrong. The FENDER is 9 doesn't mean it is divisible by 9. e.g. it could be divisible by 19. The reason is that it must have a fender of 8. you work it out why. Link to comment Share on other sites More sharing options...

Maths Seeker Posted June 12, 2006 Share Posted June 12, 2006 Has anyone got D yet? (I am in year 8 and I am not doing a booklet of some sort and am in Maths Extension) I've got 180, 360, or 720 as you can only use one set of fenders Link to comment Share on other sites More sharing options...

Maths Seeker Posted June 12, 2006 Share Posted June 12, 2006 I know all the answers so if anyone needs help, ask me. Link to comment Share on other sites More sharing options...

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