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

Info icon This preview shows pages 9–12. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 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 .
Image of page 11

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 12
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern