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Unformatted text preview: STEP Solutions 2009 Mathematics STEP 9465, 9470, 9475 The Cambridge Assessment Group is Europe's largest assessment agency and plays a leading role in researching, developing and delivering assessment across the globe. Our qualifications are delivered in over 150 countries through our three major exam boards. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate, a department of the University of Cambridge. Cambridge Assessment is a notforprofit organisation. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by the Examiners. It does not indicate the details of the discussions which took place at an Examiners’ meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. Cambridge Assessment will not enter into any discussion or correspondence in connection with this mark scheme. © UCLES 2009 More information about STEP can be found at: http://www.admissionstests.cambridgeassessment.org.uk/adt/ 2 Contents Step Mathematics (9465, 9470, 9475) Report Page STEP Mathematics I 4 STEP Mathematics II 42 STEP Mathematics III 53 3 STEP I, Solutions 2009 4 Question 1 A proper factor of an integer N is a positive integer, not 1 or N , that divides N . (i) Show that 3 2 × 5 3 has exactly 10 proper factors. It is not by accident that this question writes “ 3 2 × 5 3 ” and not “ 1125 ”: it is aiming to suggest that it is much more straightforward to think about factors of a number if we are given its prime factorisation to begin with. Also, note that the question does not ask us to multiply out the factorisations at any point. In fact, there is no need to even give the factors explicitly if you do not need to. Determining the proper factors of 3 2 × 5 3 is straightforward: any factor must be of the form 3 r × 5 s with 0 6 r 6 2 and 0 6 s 6 3, giving the factors: 3 × 5 (= 1) (this is not a proper factor) 3 × 5 1 (= 5) 3 × 5 2 (= 25) 3 × 5 3 (= 125) 3 1 × 5 (= 3) 3 1 × 5 1 (= 15) 3 1 × 5 2 (= 75) 3 1 × 5 3 (= 375) 3 2 × 5 (= 9) 3 2 × 5 1 (= 45) 3 2 × 5 2 (= 225) 3 2 × 5 3 (= 1125) (this is not a proper factor) Therefore there are 10 proper factors in total. Alternatively, we could simply note that there are 3 possible values for the power of 3 (namely 0, 1 and 2) and 4 for the power of 5 (namely 0, 1, 2 and 3), making 3 × 4 = 12 factors....
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