{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

IMC_2012_web_solutions

We have now used the digits 2 3 4 5 8 and 9 leaving 1

Info icon This preview shows pages 5–7. Sign up to view the full content.

View Full Document Right Arrow Icon
We have now used the digits 2, 3, 4, 5, 8 and 9, leaving 1, 6 and 7 . From the middle ring we have that 11 4 x w , and so 7 x w . From the second ring from the right 11 3 y x , and so 8 y x . So we need to solve the equations 7 x w and 8 y x , using 1, 6 and 7. It is easy to see that the only solution is 1 x , 7 y and 6 w . So 6 goes in the region marked *. 9. Auntie Fi’s dog Itchy has a million fleas. His anti-flea shampoo claims to leave no more than 1% of the original number of fleas after use. What is the least number of fleas that will be eradicated by the treatment? A 900 000 B 990 000 C 999 000 D 999 990 E 999 999 Solution: B Since no more than 1% of the fleas will remain, at least 99% of them will be eradicated. Now 99% of a million is 000 990 000 10 99 000 000 1 100 99 . 9 5 * 8 9 5 w * 8 v u v x y z
Image of page 5

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

View Full Document Right Arrow Icon
6 10. An ‘abundant’ number is a positive integer N , such that the sum of the factors of N (excluding N itself) is greater than N. . What is the smallest abundant number? A 5 B 6 C 10 D 12 E 15 Solution: D In the IMC, it is only necessary to check the factors of the numbers given as the options. However, to be sure that the smallest of these which is abundant, is the overall smallest abundant number, we would need to check the factors of all the positive integers in turn, until we find an abundant number. The following table gives the sum of the factors of N (excluding N itself), for 12 1 N . N 1 2 3 4 5 6 7 8 9 10 11 12 factors of N, excluding N - 1 1 1,2 1 1,2,3 1 1,2,4 1,3 1,2,5 1 1,2,3,4,6 sum of these factors 0 1 1 3 1 6 1 7 4 8 1 16 From this table we see that 12 is the smallest abundant number. Extension Problems 10.1. Which is the next smallest abundant number after 12? 10.2. Show that if n is a power of 2, and 2 n (that is, n 4, 8, 16, .. etc) then 3 n is an abundant number. 10.3 Prove that if n is an abundant number, then so too is each multiple of n . 10.4 A number, N , is said to be deficient if the sum of the divisors of N , excluding N itself, is less than N. Prove that if N is a power of 2, then N is a deficient number. 10.5 A number, N , is said to be perfect if the sum of the divisors of N , excluding N itself, is equal to N. We see from the above table that 6 is the smallest perfect number. Find the next smallest perfect number. Note: It follows from Problems 10.2 and 10.4 that there are infinitely many abundant numbers and infinitely many deficient numbers. It remains an open question as to whether there are infinitely many perfect numbers. In Euclid’s Elements (Book IX, Proposition 36) it is proved that even integers of the form ) 1 2 ( 2 1 p p , where 1 2 p is a prime number are perfect (for example, the perfect number 6 corresponds to the case where 2 p ). Euclid lived around 2300 years ago. It took almost 2000 years before the great Swiss mathematician Leonard Euler showed that, conversely, all even perfect numbers are of the form ) 1 2 ( 2 1 p p , where 1 2 p is prime.
Image of page 6
Image of page 7
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