SMC2012_sol_short.pdf - 2. B 3. D UKMT 4. B UK SENIOR...

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This solutions pamphlet outlines a solution for eachproblem on this year's paper. We have tried to give themoststraightforwardapproach,butthesolutionspresented here are not the only possible solutions.Occasionally we have added a ‘Note’ (in italics).Please share these solutions with your students.Muchofthepotentialbenefitofgrapplingwithchallengingmathematicalproblemsdependsonteachers making time for some kind of review, orfollow-up, during which students may begin to seewhat they should have done, and how many problemsthey could have solved.We hope that you and they agree that the first 15problems could, in principle, have been solved by mostcandidates; if not, please let us know.Keep these solutions secure until after the test onTUESDAY 6 NOVEMBER 2012The UKMT is a registered charity.1.3.4.5.6.7.8.9.2.10.11.12.13.14.15.21.22.23.24.25.16.17.18.19.20.SOLUTIONSUKMTUKMTUKMTUK SENIOR MATHEMATICAL CHALLENGEOrganised by theUnited Kingdom Mathematics Trustsupported byEBDBCCDCCEDEBDAABAEEDBCBB
1.EIf an odd number is written as the sum of two prime numbers then one of those primes is2, since 2 is the only even prime. However, 9 is not prime so 11 cannot be written as thesum of two primes. Note that 5 = 2 + 3; 7 = 2 + 5; 9 = 2 + 7; 10 = 3 + 7,so 11 is the onlyalternative which is not the sum of two primes.2.BThe interior angles of an equilateral triangle, square, regular pentagon are,,respectively.The sum of the angles at a point is.So.60°90°108°360°θ=360− (60+90+108) =1023.DThe cost now is.(70+4×5+6×2)=£1.02p4.BOne hundred thousand million is. So the number of stars is.102×103×106=10111011×1011=10225.CLet the required addition be, where,,,,,are single, distinct digits.To make this sum as large as possible, we need,,(the tens digits) as large as possible;sothey must be 7, 8, 9 in some order.Then we need,,as large as possible, so 4, 5, 6 insome order. Hence the largest sum is.ab+cd+efa b c d e fa c eb d f10(7+8+9) + (4+5+6) =10×24+15=2556.CIn order to be a multiple of 15, a number must be a multiple both of 3 and of 5. So itsunits digit must be 0 or 5.However, the units digit must also equal the thousands digitand this cannot be 0, so the required number is of the form. The largest such four-digit numbers are 5995, 5885, 5775. Theirdigit sums are 28, 26, 24 respectively. In orderto be a multiple of 3, the digit sum of a number must also be a multiple of 3, so 5775 isthe required number. The sum of its digits is 24.‘5aa5’7.DAdd the first and third equations:, so.Add the first two equations:, so. Substitute forandin the first equation:so.Therefore.

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Term
Fall
Professor
NoProfessor
Tags
Numerical digit, Prime number, triangle, Regular polygon

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