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FOR KIDS WHO LIKE to SOLVE PROBLEMS #2 MAGAZINE Mpower! Our STAFF ™ FOR KIDS WHO LIKE TO SOLVE PROBLEMS #2 MAGAZINE MPower! Our STAFF Editor-in-Chief Laura Gallus CONTENTS Contributing Editor Nina Dubinsky Contributing Author Andrey Polin The Three Princesses Art Direction by 1 Factors & Faulty Lights 3 KEY Grades 2 & up Grades 4 & up Amusements 5 Grades 6 & up Master Solver The Secret Hideout 8 Note from the EDITOR MChallenge 10 There is no such thing as a “math person.” However, a child’s early learning experiences directly impact their perception of their Pythagorean Means mathematics ability. When children 12 are taught mathematics through fun and engaging problems, they not only build fluency in and an intuitive understanding of mathematics, but they also develop critical problem- solving skills. MPower! was created as a resource for children who want to flex their mathematical muscles and show off their problem-solving ingenuity. We hope you enjoy the stories and problems inside. The RSM Foundation is committed to the promotion of STEM activities for K- 12 students. With the belief that Mathematics is the queen of the sciences, the Foundation advocates for ANSWERS the adoption of world -class standards and supports the Want to know if your implementation of innovative mathematical programs in a diverse context across the United States and abroad. The RSM answers are correct? Foundation strives to develop world leading intellectuals in All answers can be found online at: STEM fields, who will compete and collaborate across the www.rsmfoundation.org/mpower country. Connect with us at: ©2016 RSM Foundation. All Rights Reserved. No part of this work may be [email protected] reproduced or distributed in any form or by any means. The Three Princesses 1 The Three Princesses King Albert and King Edward wanted to unite to Prince John. However, he did not tell the their kingdoms through the marriage of King prince who is the eldest, middle, and youngest. Albert’s son, Prince John, and one of King Edward’s three daughters. King Edward invited Now, the prince intended to marry a princess he King Albert and Prince John to his castle could trust. Therefore, he decided that it must so that Prince John could choose one of the be either the eldest princess or the youngest princesses to marry. princess (he figures he can simply negate all of her responses). The prince was very clever, and It was common knowledge that the eldest he knew he could make his choice by simply princess always tells the truth, the youngest asking one of the princesses a single yes or no princess never tells the truth, and the middle question. With the three princesses standing princess, who is quite mischievous, tells the before him, he asked his question. He then truth only some of the time. King Edward made his choice. presented his daughters Princess Catherine, Princess Regina, and Princess Elizabeth What question did he ask? © 2015 RSM Foundation 2 Faulty Lights Faulty Lights Andrey Polin The holiday season was fast approaching, and Proof Miss Snow, who loved the season more than Let p be a factor of n. Then r = n/p is also factor anyone, decorated the beautiful tree in her of n. For this and each next factor we have the yard with a string of 300 lights. She turned pair (p, r). the lights on and marveled at the sight. At midnight, however, she noticed something 1. If the number of the factors is an odd number, strange happen—bulbs started blinking on it means the factor a does not pair with another or off every second. In the 1st second, every factor. However, that happens if n = a2. bulb was on; in the 2nd second every second 2. Otherwise, if n = a2, then for the factor a, the bulb, starting with bulb #2, blinked off; in pair is (a, n/a = a). So the two factors in this pair the 3rd second, every third bulb, starting from are the same and counted as one factor. Other bulb #3, blinked off or on (some bulbs that were off blinked on again—for example, bulb pairs have two different factors. It means the #6 blinked on again); in the 4th second, every number of factors is an odd number. fourth bulb, starting with bulb #4 blinked off or on; and so on. This continued for a Back to the faulty lights while, and then stopped. At what time did the In the pattern, each second corresponds to a blinking stop? How many bulbs on the string bulb number. Since there are 300 bulbs, the of lights were still on? pattern continues for 300 seconds, or 5 minutes. So, the lights stop blinking at 12:05 A.M. A little math on the side How many factors does a natural number We choose, for example, bulb #52. When does have? Every natural number, except 1, has at this bulb blink on or off? It happens in the 1st least two factors: 1 and the number itself. For second (bulb is on), in the 2nd second (bulb is example, 2 has two factors (1 and 2), 6 has th th four factors (1, 2, 3, 6), 9 has 3 factors (1, 3, 9), off), in the 4 second (bulb is on), in the 13 th 24 has 8 factors (1, 2, 3, 4, 6, 8, 12, 24). Note second (bulb is off), in the 26 second (bulb is that the numbers 2, 6, and 24 each have an on), and, finally, in the 52nd second (bulb is off). even number of factors and the number 9 has So we get 1, 2, 4, 13, 26, and 52—all the factors an odd number of factors. Without factoring, of 52. For each bulb number we get the same how can we determine if a given number has result in terms of factors. That is, the bulb will an even or an odd number of factors? only blink on or off when the number of seconds corresponds to a factor of the bulb number. We see that, for the numbers 2, 6, and 24, each factor of these numbers pairs with So, how many bulbs on the string of lights are another factor. However, the factor 3 of the still on? If the bulb number has an even number number 9 does not pair with another factor. of factors, the bulb is off. If the bulb number has an odd number of factors, the bulb is on. We A natural number n has an odd number of know that only square numbers have an odd factors if and only if n is a square number 2 2 (i.e. n = a2, where a is a natural number). number of factors. Therefore, since 1 = 1 , 4 = 2 , 9 = 32,…, 225 = 152, 256 = 162, 289 = 172, we have © 2015 Foundation RSM our answer: 17 bulbs are still on. 3 What if the string of lights has 290 bulbs? 320 bulbs? How many of the bulbs will still be on at the end of the pattern? According to our solution, a bulb will be on if the bulb number is a square number. Since both numbers are greater than or equal to 289, but less than 324 (the next square number), we know that only 17 bulbs will still be on for both of these strings of lights. Alternate blinking patterns In each second of the original pattern some bulbs blinked on or off. We can think of each second as a step of the pattern. That is, in each step of the pattern some bulbs blink on or off. What happens if we restrict when the bulbs blink on or off? The bulbs will still blink on or off when the number of seconds corresponds to a factor of the bulb number. Pattern 1 In pattern 1, bulbs blink on or off only in even seconds. So, Step 1 is the 2nd second, Step 2 is the 4th second, and so on. How many of the 300 bulbs will be on at the end of pattern 1? Solution Note that when the number of seconds is an odd number, nothing happens. So the odd-numbered bulbs will never blink on. Now we look closely at the numbers of seconds at which the steps happen: Step 1 2nd second Step 2 4th second Step 3 6th second Step 150 300th second We can see that a step number is a factor of the corresponding number of seconds. It means that, for each step, the bulb will blink on or off if the step number corresponds to a factor of the bulb number. That’s because a factor of a factor of a number is also a factor of the number. © 2015 RSM Foundation 4 Faulty Lights From our solution of the original problem, a bulb From the solution of the original problem and the will be on at the end of the new pattern if and only factorization above we can see that the answer is if the step number is a square number. We have obtained by finding all n = 2k × d, where k ≥ 0 and 150 steps and only 12 square numbers within the d is the square of an odd number (1 = 12, 9 = 32, step numbers. So, only 12 bulbs are still on at the 25 = 52, …, 169 = 132, 225 = 152, 289 = 172). So for end of the new pattern. Can you find which bulbs 300 bulbs we have: are still on? 1: 20×1=1, 2×1=2, 22×1=4, …, 28×1=256 Pattern 2 9: 20×9=9, 21×9=18, 22×9=36, …, 25×9 = 288 0 1 3 In pattern 2, bulbs blink on or off only in odd 25: 2 ×25=25, 2 ×25=50, … , 2 ×25=200 seconds.
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