June8-12Gr910Booklet-PS

CEMC at Home Grade 9/10 - Monday, June 8, 2020 Contest Day 6 Today's resource features one question from the recently released 2020 CEMC Mathematics Contests. 2020 Fryer Contest, #3 In a Dlin sequence , the first term is a positive integer and each term after the first is calculated by adding 1 to the previous term in the sequence, then doubling the result. For example, the first seven terms of the Dlin sequence with first term 4 are: 4 , 10 , 22 , 46 , 94 , 190 , 382 (a) The 5 th term in a Dlin sequence is 142. What are the 4 th and 6 th terms in the sequence? (b) Determine all possible first terms which give a Dlin sequence that includes 1406. (c) Which possible first terms from 10 to 19 inclusive produce a Dlin sequence in which all terms after the first have the same ones (units) digit? (d) Determine the number of positive integers between 1 and 2020, inclusive, that can be the third term in a Dlin sequence. More Info: Check out the CEMC at Home webpage on Monday, June 15 for a solution to the Contest Day 6 problem.

CEMC at Home Grade 9/10 - Monday, June 8, 2020 Contest Day 6 - Solution A solution to the contest problem is provided below. 2020 Fryer Contest, #3 In a Dlin sequence , the first term is a positive integer and each term after the first is calculated by adding 1 to the previous term in the sequence, then doubling the result. For example, the first seven terms of the Dlin sequence with first term 4 are: 4 , 10 , 22 , 46 , 94 , 190 , 382 (a) The 5 th term in a Dlin sequence is 142. What are the 4 th and 6 th terms in the sequence? (b) Determine all possible first terms which give a Dlin sequence that includes 1406. (c) Which possible first terms from 10 to 19 inclusive produce a Dlin sequence in which all terms after the first have the same ones (units) digit? (d) Determine the number of positive integers between 1 and 2020, inclusive, that can be the third term in a Dlin sequence. Solution: (a) If the 5 th term in a Dlin sequence is 142, then the 6 th term is (142 + 1) × 2 = 143 × 2 = 286. To determine the 4 th term in the sequence given the 5 th , we "undo" adding 1 followed by doubling the result by first dividing the 5 th term by 2 and then subtracting 1 from the result. To see this, consider that if two consecutive terms in a Dlin sequence are a followed by b , then b = ( a + 1) × 2. To determine the operations needed to find a given b (that is, to move backward in the sequence), we rearrange this equation to solve for a . b = ( a + 1) × 2 b 2 = a + 1 b 2 - 1 = a Thus if the 5 th term in the sequence is 142, then the 4 th term is 142 2 - 1 = 71 - 1 = 70. (We may check that the term following 70 is indeed (70 + 1) × 2 = 142.) (b) If the 1 st term is 1406, then clearly this is a Dlin sequence that includes 1406. If the 2 nd term in a Dlin sequence is 1406, then the 1 st term in the sequence is 1406 2 - 1 = 703 - 1 = 702. If the 3 rd term in a Dlin sequence is 1406, then the 2 nd term is 702 (as calculated in the line above) and the 1 st term in the sequence is 702 2 - 1 = 351 - 1 = 350. If the 4 th term in a Dlin sequence is 1406, then the 3 rd term is 702, the 2 nd term is 350, and the 1 st term in the sequence is 350 2 - 1 = 175 - 1 = 174.

At this point, we see that 174, 350, 702, and 1406 are possible 1 st terms which give a Dlin sequence that includes 1406. We may continue this process of working backward (dividing by 2 and subtracting 1) to deter- mine all possible 1 st terms which give a Dlin sequence that includes 1406. 1406 → 702 → 350 → 174 → 174 2 - 1 = 86 → 86 2 - 1 = 42 → 42 2 - 1 = 20 → 20 2 - 1 = 9 Attempting to continue the process beyond 9 gives 9 2 - 1 = 7 2 which is not possible since the 1 st term in a Dlin sequence must be a positive integer (and so all terms are positive integers). Thus, the possible 1 st terms which give a Dlin sequence that includes 1406 are 9, 20, 42, 86, 174, 350, 702, and 1406. (c) Each of the integers from 10 to 19 inclusive is a possible first term, and so we must determine the ones digit of each term which follows each of these ten possible first terms. If the 1 st term is 10, then the 2 nd term (10 + 1) × 2 = 22 has ones digit 2, and the 3 rd term (22 + 1) × 2 = 46 has ones digit 6. If the 1 st term is 11, then the ones digit of the 2 nd term (11 + 1) × 2 = 24 is 4, and the 3 rd term (24 + 1) × 2 = 50 has ones digit 0. Given each of the possible first terms, we list the ones digits of the 2 nd and 3 rd terms in the table below. 1 st term 10 11 12 13 14 15 16 17 18 19 Units digit of the 2 nd term 2 4 6 8 0 2 4 6 8 0 Units digit of the 3 rd term 6 0 4 8 2 6 0 4 8 2 From the table above, we see that the only ones digit which repeats itself is 8. Thus, if the 1 st term in the sequence is 18 (has ones digit 8), then the 2 nd and 3 rd terms in the sequence have ones digit 8 and so all terms will have the same ones digit, 8. Similarly, if the 1 st term in the sequence is 13 (has ones digit 3), then the 2 nd and 3 rd terms in the sequence have ones digit 8. It then follows that all further terms after the first will have ones digit 8. The 1 st terms (from 10 to 19 inclusive) which produce a Dlin sequence in which all terms after the 1 st term have the same ones digit are 13 and 18. (d) If the 1 st term in a Dlin sequence is x , then the 2 nd term is ( x + 1) × 2 = 2 x + 2, and the 3 rd term is (2 x + 2 + 1) × 2 = (2 x + 3) × 2 = 4 x + 6. For example, if x = 1 (note that this is the smallest possible 1 st term of a Dlin sequence), then the 3 rd term is 4 × 1 + 6 = 10, and if x = 2, the 3 rd term is 4 × 2 + 6 = 14. What is the largest possible value of x (the 1 st term of the sequence) which makes 4 x + 6 (the 3 rd term of the sequence) less than or equal to 2020? Setting 4 x + 6 equal to 2020 and solving, we get 4 x = 2014 and so x = 503 . 5. Since the 1 st term of the sequence must be a positive integer, the 3 rd term cannot be 2020. Similarly, solving 4 x + 6 = 2019, we get that x is not an integer and so the 3 rd term of a Dlin sequence cannot equal 2019. When 4 x + 6 = 2018, we get 4 x = 2012 and so x = 503. Thus, if a Dlin sequence has 1 st term equal to 503, then the 3 rd term of the sequence is a positive integer between 1 and 2020, namely 2018. Further, 503 is the largest possible 1 st term for which the 3 rd term has this property. Each 1 st term x will give a different 3 rd term, 4 x + 6.