1. Introduction
Number patterns are very common in Maths Olympiad questions. As most problems involve complex equations or long workings, finding out the number pattern in a question allows us to simplify the calculation process, leading to an easier method to look for the answer.
2. Basic Examples
Question 2: Find the following number of the sequence.
0, 3, 5, 8, 13, 26, 39, _ To solve this question, we must identify the sequence by coming out with a reasoning that fits the entire sequence of numbers. By observing closely, we can see that this question involves both Fibonacci and Tribonacci numbers. Fibonacci Numbers: 1, 2, 3 (1+2), 5 (2+3), 8 (3+5), 13 (8+5) Tribonacci Numbers: 1, 2, 3, 6 (1+2+3), 11 (2+3+6), 20 (3+6+11), 37 (6+11+20) 0, 3, 5, 8 (0+3+5), 13 (5+8), 26 (5+8+13), 39 (13+26), 78 (13+26+39) Answer: 78 3. Actual Competition Questions
Question 3: Twenty-five different positive integers add to 2008. What is the largest value that the least of them can have? (Source: AMC 2008 Intermediate Division Question 23)
Assuming some of the integers are consecutive, then we will get: x + (x+1) + (x+2) + ... + (x+24) = 2008 25x + 300 = 2008 25x = 2008-300 25x = 1708 1708 is not divisible by 25. Therefore, we round of 1708 to the nearest number divisible by 25 and add the subtracted amount to the 25th number of the sequence. The nearest number divisible by 25 is 1700. 1700/25 = 68 x = 68 68 + 69 + 70 + ... + 91 + (92 +8) x + (x+1) + (x+2) + ... + (x+23) + (x+24) The answer is x = 68 Answer: 68 Question 4: The symbol 3! means 3 x 2 x 1, which equals 6. Similarly, 4! means 4 x 3 x 2 x 1, which equals 24. Find the greatest prime factor of the sum 5! + 7!. (Source: MASMO 2012 Division M Question 20) To solve this question, firstly we have to identify the pattern that 5! + 7! = 5! + 5! x 6 x 7. Then, factorization will help us to find the answer. 5! + 7! = 5! + 5! x 6 x 7 = 5! (1 + 6 x 7) = 5! (43) = 5 x 4 x 3 x 2 x 1 x 43 The number 43 is a prime factor of the sum 5! + 7!. Hence, the answer is 43. Answer: 43
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