# Answers to the 2003 Pythagoras Contest 1. B) 5 2. C) 560 - 224 + 336 = 560 3. E) 7 - Prime means that ONLY 1 and itself divide the number evenly - 3 divides 12, 15 and 9. 2 divides 12 and 4. 4. D) 5 - 25 divided by 5 is 5 5. B) 5 - 24 can be divided evenly by each other number. 6. A) 900 - (91 x 9) = 819, (9 x 9) = 81, 819 + 81 = 900 - Also, but associativity, (91 x 9) + (9 x 9) = (91 + 9) x 9 = 100 x 9 7. D) 14 - (12 + 14 + 16) = 42. 42 divided by 3 is 14 8. E) 70 9. E) 500 10. C) 10 - If it's a pile, then blocks can't float in the air, the top two layer need blocks under them. There's 1 in the top layer, 4 in the second and 5 on the bottom. 1 + 4 + 5 = 10 11. D) 4 divided by 4 - 4 divided by 4 is 1 12. D) 100 - There are 5 x 5 = 25 fours. 25 x 4 = 100 13. C) 30 - A pentagon has 5 sides, a Hexagon has 6, 5 x 6 = 30 14. B) 20 - 8 quarters is $2.00, 10 dimes is $1.00. So we need enough nickels to make $1.00, or 20 15. C) 5 - 1 + 2 + 3 + 4 + 5 = 15, which is a multiple of 5 16. A) 4 484 - 4 000 + 440 + 44 = 4 484 17. C) 1/8 - It's less than a quarter. I think it's 1/8 18. E) 0 - 100 x 111 = 11 100, they just want the tens digit 19. C) 1 - 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16. 4 x 4 = 16. And they want N divided by 4, so 1 20. D) 10 - Work backwards. One half of 10 is 5. 5 more than that number is 5. 21. C) April 8 - The fourth day before the third day after is the day before. 22. B) 5 3/4 (five and three quarters, or 5 hours, 45 minutes) - Let's write 2:30 PM as 14:30, 14:30 - 8:45 = 5:45 hours. It stopped at least 5:15 hors ago, because it could've stopped days (or even years ago). 23. E) (M + 2) + (A + 2) - Their ages together are M + A, in two years they will be M + 2 + A + 2 24. A) 1/36 - There are 36 different ways that 2 dice can come up (a 3 on the first and a 4 on the second is different from a 4 on the first and a 3 on the second, so we have 6 x 6 = 36 possibilities. Each one is equally likely). Only 1 of them, 6 and 6, gives 12. 25. D) 650 - Notice that (4 + 8 + 12 + ... + 100) is twice (2 + 4 + 6 + ... + 50), and then we take it away, leaving only 650. 26. E) 69 - the second = the first + 1 - third = second + 1 = first + 2 - fourth = third + 1 = first + 3 - fifth = fourth + 1 = first + 4 - sixth = fifth + 1 = first + 5 - thus: 399 cm - = first + first + 1 + first + 2 first + 3 + first + 4 + first + 5 - = (6 x first) + 15 cm - so 384 cm = (6 x first), or the first is 64 cm - thus the sixth, and longest is first + 5 = 64 + 5 = 69 cm 27. C) 102 cm - Each triangle adds 1 cm to the top or bottom, but the side stay the same length (they do change angles though). So 100 triangles make for 100cm on the top or bottom and one on each ends. 28. C) 3 - Since 4 quarters are $1.00, she had to have 3 or less. - She can't have 1, because even with 5 dimes, you can only get to 75 cents - With 2 you have 50 cents and 4 coins left. 3 have to be dimes to get to 80, then you have one coin to make 2 cents, which doesn't work. - So she has 3 quarters, a nickel and 2 pennies. 29. B) 0 - Hmm, anything with one digit times 9 follow a pattern. 2 x 9 = 18, the first digit is the number minus 1, and the two ad d up to 9. - E has to be 1, since otherwise the product MORE would have 1 more digit than EROM. (E can't be 0, because then M would also be 0) - That means that M has to be 9. - 1RO9 x 9 = 9OR1. Let's look at R, it has to be zero or else it would have two digits and interfere with our E, M combination. - So if 10z9 x 9 = 9z01, that's enough that we can just use trial and error, giving z = 8 - and sure enough 1089 x 9 = 9801 30. B) 9 cm - Hmm, well, look at the smallest squares; I'm going to guess they are 1 cm by 1 cm. - Then the next biggest is 2 cm, the next is 3 cm, So the largest is 2 + 3 = 5 cm. The last two are the same size, and their sides add up to 5 + 3 = 8, so they are 8. Ok, so that would mean that the rectangle was 8 cm by 9 cm... Except that it is really 72 cm by 81 cm, or 9 times my guess, so all the squares are 9 times what I thought. 31. A) 12.8 - 12.80 = 12.80 32. A) 2 468 642 33. B) 1/5 - 7 + 13 = 20, 20 /100 = 1 / 5 (Don't pick up broken glass) 34. D) - A greegree has the dot in all three shapes at the same time 35. D) 9 - Each fraction is 1 / 3, so 27 of them is 27 / 3 or 9 36. D) 153 - 1 cubed = 1, 5 cubed = 125, 3 cubed = 27, 1 + 125 + 27 = 153 37. D) 7 x 7 x 7 - 7 x 7 x 7 = 343 38. A) 17 - 2, 5 and 10 work. I think that they mean between 1 and 20, but not 1 or 20 39. A) 3 - if 12 B = 18 G, then 2 B = 3 G. Since he has enough left to feed 6 girls, he has enough to feed 4 boys. But he's a boy, and I assumed he's coming, so that would be 3 more that he can invite. 40. C) 6 - If both of the first numbers are the same, then it's a multiple of the regular Fibonacci sequence, so - 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 is 143, or a third of the number she sum she has, so her numbers are 3 times mine, thus her third is 6 41. C) 8 - plus 4, minus 2, plus 4.... 42. D) N and A - Melissa is at the end (or else she'd have someone else beside her), beside Nicole. - Next is Andrea, and Carol is at the end. 43. C) 90 - A, b and the 90 degree angle add up to 180, so they must add up to 90 on their own. 44. A) A = p/2 + i - 1 - Rule them out one by one. - The square has p = 4 perimeter points, i = 0 inside points, and A = area of 1 - So it rules out B, C, D, and E. That leaves just A - let's check with one other shape (the lower left corner) has area 5, p = 8, i = 2 - 5 = 8 / 2 + 2 - 1 = 4 + 2 -1, so it looks like A 45. B) 355/113 - 1/7 is about .143 - 21/7 is 3, so 20/7 is too small - so 22/7 is about 3.14, very close, and 23/7 is too big - the problem is 355/113 - You have to do long division on - 22/7 = 3.1428571428571428571428571428571 - and 355/113 = 3.1415929203539823008849557522124 - I used a calculator, but you only have to go to 5 digits to see that 355/113 is closer 46. E) 1% - 40 cm is 1 percent of 40 m 47. D) 432 - You can either find values for M, N, O and P or see that - M x N x O x P = ( M x N ) x ( P x O ) = 12 x 36 = 432 48. B) 30.5 - By the Pythagoran Theorem, the radius is 10 cm, so the circle's area is about 314. We just want one quarter, or about 78.5, and subtract the rectangle, which is 48, so 30.5 49. C) 3 - I don't care about the colours, 7 + 6 + 5 = 18 balls, if we count every bag twice, or 9 left total. So he took out 3 balls (1 from II and 2 from I, actually) 50. E) impossible - There will always be an odd number of cups "right side up" -- he can't change that.