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Mathematics STEP Dr Chris Warner College Teaching Officer Director of Studies in Mathematics Educational Consultant Robinson College Why STEP? • • • • • Reveals greater depth of understanding Measures ability to think “outside the box” Tests lateral thinking Tests ability to apply knowledge A level is not (we believe) a good (enough) preparation for mathematics in top universities Robinson College STEP resources • Cambridge Assessment’s Admissions Testing Service web site: − http://www.admissionstestingservice.org/for-testtakers/step/about-step/ • OpenBook publishers web site: − Advanced Problems in Mathematics: Preparing for University http://www.openbookpublishers.com/product/342/advanced -problems-in-mathematics--preparing-for-university ‒ purchasable book and downloadable PDF by Stephen Siklos ‒ hints and full solutions ‒ a good starting point for preparation Robinson College STEP resources • Mathematics Department web site: ‒ NRICH: STEP preparation archive https://nrich.maths.org/10622 ‒ Meikleriggs Mathematics http://meikleriggs.org.uk/ ‒ complete solutions to STEP papers (latest 2010) ‒ STEP support programme: Online course to help prepare for STEP https://maths.org/step/ Robinson College STEP administration • Administered by Cambridge Assessment ‒ ‒ ‒ ‒ STEP = “Sixth Term Examination Paper” well-established Mathematics university admissions test managed by a specialist team whose remit is assessments relating specifically to university entrance dedicated website: http://www.admissionstestingservice.org/for-testtakers/step/about-step/ Robinson College STEP administration • Administered by Cambridge Assessment ‒ website provides information on ‒ method of entry ‒ Dates ‒ Syllabus ‒ downloadable past papers Robinson College STEP specification • Test summary ‒ ‒ ‒ ‒ ‒ ‒ ‒ three papers candidates typically sit 2 papers pen and paper test 3 hours six questions from 13 no calculators taken in late June Robinson College STEP papers and offers • Three STEP mathematics papers I, II, and III ‒ thirteen questions: ‒ 8 pure ‒ 3 mechanics ‒ 2 statistics and probability ‒ assessed on answers to at most 6 questions ‒ five grades (from highest to lowest) S, 1, 2, 3, and U Robinson College STEP papers and offers • Cambridge offer usually includes grades in two papers ‒ not taking the full Further Mathematics A level or equivalent: Papers I and II; ‒ offer is usually 1,1; sometimes 1,2 ‒ taking Further Mathematics A-level or equivalent: Papers II and III ‒ offer is usually1,1 Robinson College STEP papers and offers • Paper I ‒ ‒ ‒ ‒ ‒ syllabus based on single subject A level pure mathematics content slightly more than common core mechanics and probability and statistics sections equivalent to two or three A-level modules intended for candidates not taking Further Mathematics questions easier than paper II • Paper II ‒ ‒ same content as Paper I harder questions Robinson College STEP papers and offers • Paper III ‒ ‒ syllabus based on further maths A level questions same level of difficulty as paper II Robinson College Preparing for STEP • Can candidates prepare without help? – we make adjustments – arguably these may not go far enough – students need to be able to cope when they start – some “training” is essential – the Cambridge Mathematics Department runs an online STEP support programme – more likely to relax on STEP offers in August for candidates who have no input from their school Robinson College Preparing for STEP • How teachers can help – STEP is aimed at the top few students so use as differentiated material – it can be hard to do this – … but it is not impossible, particularly for further mathematics groups – Remind students that training has a huge effect – as teachers would expect – so start early and practice regularly and frequently – perseverence is good preparation for University and life! Robinson College Preparing for STEP • How teachers can help – reassure students that STEP is indeed “very hard” (as quite a few teachers have told me) – questions are hard compared to A level – there is a commonality of style – encourage students to try the online STEP support programme – reminder: this is aimed to “close the gap” for candidates who do not receive help in school; who are in schools with few applicants for Cambridge; who trigger social deprivation “flags” Robinson College A STEP paper I question A positive integer with 2𝑛 digits (the first of which must not be 0) is called a balanced number if the sum of the first 𝑛 digits equals the sum of the last 𝑛 digits. For example 1634 is a 4-digit balanced number, but 123401 is not a balanced number. (i) Show that seventy 4-digit balanced numbers can be made using the digits 0, 1, 2, 3, and 4. 1 (ii)Show that 𝑘 𝑘 + 1 4𝑘 + 5 4-digit balanced numbers can 6 be made using the digits 0 to 𝑘. You may use the identity 𝑛 2 𝑟=0 𝑟 1 6 = 𝑛 𝑛 + 1 2𝑛 + 1 . Robinson College What skills are being tested? • Being systematic and methodical • Spotting patterns • Creating formulas to count the number of possible sums and grouping the terms appropriately • Using the formula for sum of an arithmetic sequence (in formula booklet) • Using given formula for sum of first 𝑛 square numbers Robinson College One possible solution Consider combinations of first two digits and present in a table: 00 01 02 03 04 10 11 12 13 14 20 21 22 23 24 30 31 32 33 34 40 41 42 43 44 Robinson College One possible solution The first half of the four digit number cannot start with a 0 whereas the second half can start with a 0. The number of possibilities for each sum are therefore: Sum: 1 First 1 half Second 2 half 2 2 3 3 4 4 5 4 6 3 7 2 8 1 3 4 5 4 3 2 1 Robinson College One possible solution Thus the number of 4-digit balanced numbers is 1×2 + 2×3 + 3×4 + 4×5 + 4 × 4 + 3 × 3 + 2 × 2 + 1 × 1 = 70 There are two qualitatively different types of term. The first four take the form of a sum of 𝑟(𝑟 + 1) for 𝑟 = 1 to 4 and the second four take the form of a sum of 𝑟 2 for 𝑟 = 1 to 4. Robinson College One possible solution For four digit balanced numbers made using the digits 0 to 𝑘, the sum of terms generalises straightforwardly to: 𝑘 𝑟=1 𝑟 𝑟+1 + 𝑘 2 𝑟 𝑟=1 = 𝑘 2 (2𝑟 𝑟=1 + 𝑟) 1 1 = 2 × 𝑘 𝑘 + 1 2𝑘 + 1 + 𝑘(𝑘 + 1) 6 2 1 1 = 𝑘 𝑘 + 1 4𝑘 + 2 + 3 = 𝑘 𝑘 + 1 4𝑘 + 5 6 6 Robinson College