Solutions. End of Year Competition 12/10/2016

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1 12/10/2016 End of Year Competition Solutions 1. Sasheer has $93, Aidy has $5 more than Cecily, and Cecily has one third as much as Sasheer. How much money does Aidy have? Solution. $ = $ Leonardo had to pass 7 guarded gates to get into an orchard. He collected a certain number of apples from the orchard. On leaving he gave the first guard half of the apples he had and 1 apple more. He gave the second guard half of his remaining apples, and one apple more. He did the same to each of the five remaining guards, and left the orchard with 1 apple. How many apples did he gather in the orchard? Hint: It s a lot of apples. Solution. It might be easiest to work this problem backwards. Going backwards, gets first one apple from each guard and then doubles the number of apples he has. He ended with one apple, so before guard 7 he had (1+1) 2 = 4 apples. Before guard 6, he had (4+1) 2 = 10 apples, Before guard 5, 22 apples, before guard 4, 46 apples, before guard 3, 94 apples, before guard apples, before guard 1, 382 apples. 3. The great Pharaoh Nobonosehim III had a pyramid built whose base was a square of sides 100 feet long. It lateral faces were equilateral triangles. How tall was the pyramid? (The answer could contain a square root)

2 MCFAU/2016/12/10 2 Solution. From the top T of the pyramid draw a perpendicular to the base, intersecting the base at the center C. Draw the segment joining C to A, one of the vertices of the base. Triangle ACT is a right triangle with (in feet) AT = 100, by the theorem of Pythagoras (2 AC ) 2 = so that AC = Applying the theorem of Pythagoras once more we get so the answer is 50 2 feet. CT 2 = (50 2) 2 = 5000, 4. The average age of 24 kids in Ms. Krebapple s karate class is 11 years. One day one kid did not show up and the average age of the remaining 23 kids became 10 years. How old is the missing kid? Solution. On a usual day, the sum of ages of the kids is = 264. On the day of the missing kid, they only add up to 230. So the age of the missing kid is = 34 years. Note: It was a very old kid. 5. Kirsten runs twice as fast as she walks. Last Monday she went to school, she walked for twice the time that she ran. It took her 20 minutes to get to school. On Tuesday she ran for twice the time that she walked. How long did it take her to get to school on Tuesday? Solution. Suppose d is the distance to the school. Since it doesn t seem to matter what it is we can make any number we wish (or keep it indicated). I ll keep it indicated; that is, keep calling it d. Since she runs twice as fast as she walks and since on Monday she walked twice the time she ran, it means she covered half the distance running, half of it walking. Also, of the 20 minutes she spent 20/3 running and 40/3 walking. If her walking speed is w, this means that half the distance to her school,namely d/2 equals (40/3)w so d = (80/3)w. The next day she ran for twice the time that she walked, so she covered running four times the distance she walked. So she ran for (4/5)d and walked for (1/5)d Dividing by the velocities (respectively 2w, w) and adding gives us the time: 4d 10w + d 5w = x, x being the time we have to figure out. Multiplying this last equation by 10w we get 6d = 10wx. Substituting for the value of d found before, namely d = (80/3)w gives w = 10wx, canceling w and solving we get x = 16. The answer is 16 minutes. 6. What are the last two digits of ?

3 MCFAU/2016/12/10 3 Solution. The last two digits of a number are the remainder of dividing the number by 100 and one can show that the last two digits of sums or products are the corresponding sums and products of the last two digits. One solution is as follows: If we look at powers of 7 and consider only the last two digits we have 7 0 = = = (3) = (...)01 and the pattern repeats. So the last 2 digits follow the pattern 1, 7, 49, 43, 1, 7, 49, 43,... Now = 100, so the sum of the last two digits of every group of four consecutive powers beginning with 1 are 00. So if we add all the way to 7 99, the last two digits of the sum so far will be 00; then we add = 57. The answer is At dawn, two tourists simultaneously left points A and B along the same path with each heading toward the other point. They passed each other at noon, without stopping. The first came to point B at 4PM and the second cam to point A at 9PM. If each walked at a constant rate, at what time did the sun rise that day? Solution. Let us say it takes them x hours before they meet at noon. So tourist A walked a total of x + 4 hours, the second one a total of x + 9 hours. This means that the first tourist is (x + 9)/(x + 4) times faster than the second one. Or if v A, v B are the respective speeds, then v A /v B = (x + 9)/(x + 4). Now the first tourist did in 4 hours the trip from the common meeting point to B, which took the second tourist x hours to complete. This tells us that v A /v B = x/4. Equating x + 9 x + 4 = 4 x, which solves first to 4x + 36 = x 2 + 4x, then to x = 6. The answer is 12 6 = 6 AM. 8. If n is a positive integer, then we write n!, call it the factorial of n, for the product of all numbers from 1 to n. For example: 1! = 1, 2! = 1 2 = 2, 3! = = 6, 4! = = 24, 5! = = 120, 6! = = 720, 7! = = 5040, etc. They grow very fast, for example 10! = 3, 628, 800. What is the largest number of times that 3 will divide exactly into 50! = ? (For example 3 divides exactly once into 3! = 6, exactly 4 times into 9! = 362, 880.) Solution. Every third factor in 50! is a multiple of 3; there are 50/3 = 16 such factors. But every ninth factor is a multiple of 9, increasing by one the number of times 3 divides 50!. There are 50/9 = 5 such factors. Finally, there is one multiple of 27, namely 27. The answer is = 22 times. 9. A circle of center O and radius 5 contains three smaller circles; two are tangent to each other at O and a third, smaller one is tangent to the other two; all three are tangent to the outer circle, as shown below. Find the radius of the smaller (shaded) circle.

4 MCFAU/2016/12/10 4 Solution. Let A, be the center of one the circles touching at O, C the center of the smallest circle. Let R = 5 be the radius of the outer circle, then each of the two circles touching at O has radius R/2; let r be the radius of the little circle; the radius we have to find. Triangle AOB is a right triangle with hypotenuse AB of length R/2 + r, and side OA, of length R/2, and side OB of length R r. By the theorem of Pythagoras, we get ( ) 2 R 2 + r = ( ) 2 R + (R r) 2. 2 Expanding, canceling what can be canceled, etc., we get 3Rr = R 2, hence r = R/3. The answer is 5/ ABCD is a parallelogram. A segment drawn from vertex E intersects diagonal AC at E, side AB at F and the continuation of side CB at G. If DF = 6 and F G = 3, what is the length of DE? Similarity of triangles can play a role here.

5 MCFAU/2016/12/10 5 Solution. One sees that F BG F AD and ADE CGE. From the first similarity of triangles and the given information we get that since AD = BC we see that AD BG = DF = 2, so AD = 2 BG ; F G CG = CB + BG = 3 BG = 3 2 CB = 3 2 AD. From the second pair of similar triangles we now get that thus ED = 2 EG. On the other hand 3 ED EG = AD CG = 2 3, ED + EG = DF + F G = 9, from the last two equations involving ED we can solve for ED = DE to get ED = 18 5 or ED = Find the sum of the angles MAN + MBN + MCN + MDN.

6 MCFAU/2016/12/10 6 Solution. If we shift angle MAN in a parallel way 3 units to the right so the upper vertex is at D, then angle MBN by 2 units to the right so vertex B coincides with D, finally angle MCN one unit to the right, the picture looks like We see that the summ of the angles works out to A boat goes downriver from town A to town B in 3 days and upriver from town B to town A in 5 days. How long will it take a raft to float from town A to town B. Assume the water velocity is always the same, and the boat has the same speed relative to the water in both directions. Solution. Let us suppose for example that the speed of the water is 1 (could be 1 mile per day, one league per day, one yard per day, or whatever per day). If v is the speed of the boat with respect to the water, than its downriver speed is v + 1 so the distance covered in 3 days is 3(v + 1). Upriver its speed is v 1, so it covers 5(v 1) in 5 days. Since the distance from A to B is the same as from B to A, we must have 3(v + 1) = 5(v 1) or 3v + 3 = 5v 5, from which v = 4. The distance from A to B is 3(v + 1) = 15 (or 5(v 1) also equal to 15); the raft floating a speed 1 would thus take 15 days to get from A to B.