Registration is now open for Elementary MathFest Spring Tournament. The tournament is open to 4th and 5th graders and will be held on March 18th, 9:00 am to 11:30 am. The cost for the tournament is $20 if paid before competition day. Please register your student at www.waltonmathteam.com and join us at Walton High School on March 18th! Please direct any questions to Laura Speer and laura.speer@cobbk12.org.
Laura Speer Walton High School Mathematics Department
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1) If a kindergarten teacher places her children 4 on a bench, there will be 3 children who will not have a place. However, if 5 children are placed on each bench, there will be 2 empty spaces.
What is the smallest number of children the class could have? 2) If the digits A, B, and C are added, the sum is the twodigit number AB as shown below. What is the value of C? A B + C A B 3) When I open my mathematics book, there are two pages which face me and the product of the two page numbers is 1806. What are the two page numbers? Solutions: 1) Since the number of children is 3 more than a multiple of 4, that number could be 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, . . . Since the number of children is 2 less than a multiple of 5, the number could be 3, 8, 13, 18, 23, 28, 33, 28, 43, . . . The numbers satisfying both conditions are 23, 43, 63, 83, and so forth. The smallest of these numbers is 23. Thus, there are 23 children in the class. 2) In the units column, notice that the sum of A, B, and C ends in B. Then A + C = 10. Since A is also the tens digit of the sum, A must be 1. Therefore C = 9. 3) If page numbers are in the 40s, then the product is greater than 1,600. If the page numbers are in the 50s, then the product is greater than 2,500. Clearly the page numbers must be in the 40s. Since the two page numbers are consecutive numbers, the units digits must be 2 and 3, or 7 and 8. Try 42 and 43. They work! (Page numbers 47 and 48 don't work). 1) A bag contains 500 beads, each of the same size, but in 5 different colors. Suppose there are 100 beads of each color, and I am blindfolded. What is the fewest number of beads I must pick to
be absolutely sure there are 5 beads of the same color among the beads I have picked blindfolded? 2) A number has a remainder of 1 when divided by 4, a remainder of 2 when divided by 5, and a remainder of 3 when divided by 6. What is the smallest number that has the above properties? 3) A total of fifteen pennies are put into four piles so that each pile has a different number of pennies. What is the smallest possible number of pennies that could be in the largest pile? Solutions: 1) Suppose I select 5 beads blindfolded. The 5 beads could contain1 bead of each color. Suppose I select 5 beads three more times with the same result. I will then have a total 20 beads consisting of 4 beads of each color. The selection of the 21st bead guarantees that there are now at least 5 beads of the same color among the 21 beads. 2) Let N be the required number. Notice that when N is divided by 4, 5, or 6, the remainder is 3 less than the divisor in each case. If N is increased by 3, this new number N + 3 will be divisible by 4, 5, and 6. The smallest number that N + 3 can be is the least common multiple of 4, 5, and 6 which is 60. Therefore the required number is 57. 3) Method 1: The different ways 15 pennies can be distributed into 4 piles are (1, 2, 3, 9), (1, 2, 4, 8), (1, 2, 5, 7), (1, 3, 4, 7), (1, 3, 5, 6), (2, 3, 4, 6). Thus, the number of pennies in the largest pile of each distribution may be 6, 7, 8, or 9. The smallest of these is 6. Method 2: To find the smallest possible number of pennies in the largest pile, make the sum of the numbers of pennies in the other three piles as large as possible. This occurs when the first three piles contain 1, 3, and 5 pennies, or 2, 3, and 4 pennies. In either case the fourth pile will contain 6 pennies. 1) There are many numbers that divide 109 with a remainder of 4. List all twodigit numbers that have that property.
2) If a number ends in zeros, the zeros are called terminal zeros. For example, 520,000 has four terminal zeros, but 502,000 has just three terminal zeros. Let N equal the product of all natural numbers from 1 through 20: N = 1 x 2 x 3 x 4 x . . . x 20. How many terminal zeros will N have when it is written in standard form? 3) The XYZ club collected a total of $1.21 from its members with each member contributing the same amount. If each member paid for his or her share with 3 coins, how many nickels were contributed? Solutions 1) If 4 is subtracted from 109, the result is 105. Then each of the twodigit numbers that will divide 109 with a remainder of 4 will divide 105 exactly. Thus, the problem is equivalent to finding all twodigit divisors of 105. Since the prime factors of 105 are 3, 5, and 7, the divisors are 3 x 5, 3 x 7, and 5 x 7, or 15, 21, and 35. 2) Multiplication by 10 produces an additional terminal zero when a product is written in standard form. If 1 x 2 x 3 x . . . x 20 is written as a product of prime factors, it will contain four 5s and more than four 2s among many factors. Then part of the product can be written as: (5 x 2) (5 x 2) (5 x 2) (5 x 2) which can also be represented as 10 x 10 x 10 x 10. Therefore there are four terminal zeros in the product. 3) Since 121 = 11 x 11, the club has 11 members and each contributed $0.11. Each $0.11 share was paid in 3 coins which had to be 2 nickels and 1 penny. Then the 11 members contributed a total of 22 nickels. 
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