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In other words j n = j n 1 j n 2 this shows that j n

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Unformatted text preview: In other words J n = J n- 1 + J n- 2 . This shows that J n is the n + 1 Fibonacci number. Hence J 1 = 1 , J 2 = 2 , J 3 = 3 , J 4 = 5 , J 5 = 8 , J 6 = 13 , J 7 = 21 , J 8 = 34 , J 9 = 55 , J 10 = 89 , J 11 = 144 . 9 The correct answer is (f) . Solution of problem 1.3: The biggest Fibonacci number which is less than 175 is 144. So 175 = 144 + 31. The biggest Fibonacci number which is less than 31 is 21. So 175 = 144 + 21 + 10. The biggest Fibonacci number less than 10 is 8, and so 175 = 144 + 21 + 8 + 2 which are all Fibonacci numbers. The sum of the smallest and biggest of those is 2 + 144 = 146. The correct answer is (b) . Solution of problem 1.4: Clearly p must be a two digit prime number since if it is a one digit prime, then the sum of its digits can not be 16. Since the two digits of p add to 16, each of these digits must be at least 7. So the digits of p are among the numbers 7, 8, 9. But we can not have 8 as a digit of p . Indeed 8 can not be the last digit of p since then p will be even and hence not prime. Similarly 8 can not be the first digit of p since then 8 will be the last digit of q , and so q will be even and not prime. The only option then is for the digits of p to be 7 and 9. By inspection we see that both 79 and 97 are prime. So we have that either p = 79 or p = 97. In either case the product of the digits of p is 63. The correct answer is (c) . Solution of problem 1.5: First we find the prime decomposition of 3240. Since 3240 is even we can divide it by 2: 3240 = 2 · 1620 = 2 · 2 · 810 = 2 · 2 · 2 · 405. 405 is divisible by 5 so we have 3240 = 2 3 · 405 = 2 3 · 5 · 81 = 2 3 · 3 4 · 5. Therefore the correct answer is (d) . Solution of problem 1.6: By the formula for the geometric progression we know that 1 + 3 + 3 2 + ··· + 3 15 = 3 16- 1 3- 1 = 3 16- 1 2 . We have that 3 3 = 27 = 4 · 7- 1, i.e. 3 3 =- 1 mod 7 . 10 This gives 3 16 = (3 3 ) 5 · 3 = (- 1) 5 · 3 mod 7 =- 3 mod 7 . In particular 3 16- 1 =- 3- 1 =- 4 mod 7 , and so (3 16- 1) / 2 =- 4 / 2 =- 2 = 5 mod 7 . The correct answer is (d) . Solution of problem 1.7: Denote the missing digit by x . Then the check- sum formula for the barcode 0 78001 26 05 2 reads 3 · 0 + 7 + 3 · 8 + 0 + 3 · 0 + 1 + 3 · 2 + 6 + 3 x + 0 + 3 · 5 + 2 = 0 mod 10 . Simplifying we get 3 x + 1 = 0 mod 10 , or equivalently 3 x = 9 mod 10 . Thus x = 3 and so correct choice is (b) . Solution of problem 1.8: (i) Suppose Q := 2 √ 5- 1 is rational. Then √ 5 = Q + 1 3 . Since 1 and 3 are rational numbers, and since arithmetic operations on natural numbers produce rational numbers, we conclude that √ 5 is rational. This is a contradiction, since we have shown that √ 5 is irrational. Thus 2 √ 5- 1 is irrational ....
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In other words J n = J n 1 J n 2 This shows that J n is the...

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