SMC2012_web_solutions

Occasionally we have added a note in italics please

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Occasionally we have added a ‘Note’ (in italics). Please share these solutions with your students. Much of the potential benefit of grappling with challenging mathematical problems depends on teachers making time for some kind of review, or follow-up, during which students may begin to see what they should have done, and how many problems they could have solved. We hope that you and they agree that the first 15 problems could, in principle, have been solved by most candidates; if not, please let us know. Keep these solutions secure until after the test on TUESDAY 6 NOVEMBER 2012 The 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. SOLUTIONS UKMT UKMT UKMT UK SENIOR MATHEMATICAL CHALLENGE Organised by the United Kingdom Mathematics Trust supported by E B D B C C D C C E D E B D A A B A E E D B C B B
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1. E If an odd number is written as the sum of two prime numbers then one of those primes is 2, since 2 is the only even prime. However, 9 is not prime so 11 cannot be written as the sum of two primes. Note that 5 = 2 + 3; 7 = 2 + 5; 9 = 2 + 7; 10 = 3 + 7, so 11 is the only alternative which is not the sum of two primes. 2. B The 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 ) = 102 3. D The cost now is . ( 70 + 4 × 5 + 6 × 2 ) = £1 . 02 p 4. B One hundred thousand million is . So the number of stars is . 10 2 × 10 3 × 10 6 = 10 11 10 11 × 10 11 = 10 22 5. C Let 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; so they must be 7, 8, 9 in some order. Then we need , , as large as possible, so 4, 5, 6 in some order. Hence the largest sum is . ab + cd + ef a b c d e f a c e b d f 10 ( 7 + 8 + 9 ) + ( 4 + 5 + 6 ) = 10 × 24 + 15 = 255 6. C In order to be a multiple of 15, a number must be a multiple both of 3 and of 5. So its units digit must be 0 or 5. However, the units digit must also equal the thousands digit and this cannot be 0, so the required number is of the form . The largest such four- digit numbers are 5995, 5885, 5775. Their digit sums are 28, 26, 24 respectively. In order to be a multiple of 3, the digit sum of a number must also be a multiple of 3, so 5775 is the required number. The sum of its digits is 24. ‘5 aa 5’ 7. D Add the first and third equations: , so . Add the first two equations: , so . Substitute for and in the first equation: so . Therefore . 2 x = 4 x = 2 2 x + 2 y = 3 y = - 1 2 x y 2 + ( - 1 2 ) + z = 1 z = - 1 2 xyz = 2 × ( - 1 2 ) × ( - 1 2 ) = 1 2 8. C If an equilateral triangle is split into a number of smaller identical equilateral triangles then there must be one small triangle in the top row, three small triangles in the row below, five small triangles in the row below that and so on. So the total number of small triangles is 4 or 9 or 16 etc. These are all squares and it is left to the reader to prove that the sum of the first odd numbers is .
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