This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full Document
 Spring '08
 schneps
 Calculus, Decimal, Natural number, Prime number

Click to edit the document details