MATH
IMC_2012_web_solutions

# 11 check which of the numbers 1 11 111 and 1111 if

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1.1 Check which of the numbers 1, 11, 111 and 1111, if any, are prime. 1.2 Show that a positive integer all of whose digits are 1s, and which has an even number of digits, is not prime. 1.3 Show that a positive integer all of whose digits are 1s, and which has a number of digits which is a multiple of 3, is not prime. 1.4 Show that a positive integer all of whose digits are 1s, and which has a number of digits which is not a prime number, is itself not a prime number. 1.5 It follows from 1.4 that a number all of whose digits are 1s can be prime only if it has a prime number of these digits. However numbers of this form need not be prime. Thus 11 with 2 digits is prime, but 111 with 3 digits is not. Determine whether 11111, with 5 digits, is prime. Without a computer quite a lot of arithmetic is needed to answer question 1.5. As numbers get larger it becomes more and more impractical to test by hand whether they are prime. Using a computer we can see that 1111111, 11111111111, 1111111111111 and 11111111111111111 with 7, 11, 13 and 17 digits, respectively, are not prime. In fact, 4649 239 1111111 , 513239 21649 1 1111111111 , 265371653 79 53 111 1111111111 and 5363222357 2071723 1111111 1111111111 , where the given factors are all prime numbers. The next largest example after 11 of a number all of whose digits are 1s and which is prime is 1111111111111111111 with 19 digits. The next largest has 23 digits, and the next largest after that has 317 digits. It is not known whether there are infinitely many prime numbers of this form. We have taken this information from the book The Penguin Dictionary of Curious and Interesting Numbers by David Wells, Penguin Books, 1986.

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3 2. Three positive integers are all different. Their sum is 7. What is their product? A 12 B 10 C 9 D 8 E 5 Solution: D It can be seen that 7 4 2 1 and 8 4 2 1 , so assuming that there is just one solution, the answer must be 8. In the context of the IMC, that is enough, but if you are asked to give a full solution, you need to give an argument to show there are no other possibilities. This is not difficult. For suppose a , b and c are three different positive integers with sum 7, and that c b a . If 2 a , then 3 b and 4 c , and so 9 c b a . So we must have that 1 a . It follows that 6 c b . If 3 b then 4 c and hence 7 4 3 c b . So 2 b . Since 1 a and 2 b , it follows that 4 c . 3. An equilateral triangle, a square and a pentagon all have the same side length. The triangle is drawn on and above the top edge of the square and the pentagon is drawn on and below the bottom edge of the square. What is the sum of the interior angles of the resulting polygon? A 0 180 10 B 0 180 9 C 0 180 8 D 0 180 7 E 0 180 6 Solution: E The sum of the interior angles of the polygon is the sum of the angles in the triangle, the square and the pentagon. The sum of the interior angles of the triangle is 0 180 , and the sum of the angles of the square is 0 0 180 2 360 , and the sum of the angles of the pentagon is 0 0 180 3 540 .
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• Spring '13
• MRR
• Math, Prime number, triangle, Divisor, perfect number

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