MOEMS posts a problem of the month each month at http://www.moems.org/zinger.htm
Here are some challenging word problems that are similar to what 4th and 5th graders experience at a Math Olympiad. Math team members can use these for extra practice and everyone else can use them to help extend your problem solving skills. Solutions are at the bottom of this post.
1) Suppose today is Tuesday. What day of the week will it be 100 days from now?
2) The four-digit numeral 3AA1 is divisible by 9. What does A represent?
3) A purse contains 4 pennies, 2 nickels, 1 dime, and 1 quarter. Different values can be obtained by using one or more coins in the same purse. How many different values can be obtained?
1) Every 7 days from "today" will be Tuesday. Since 98 is a multiple of 7, the 98th day from today will be Tuesday. Then the 100th day from today will be Thursday.
2) If the number is divisible by 9, then the sum of its digits is divisible by 9. The digit sum is 3 + A + A + 1 = 4 + 2A. The digit sum cannot by 9, otherwise A = 2 1/2. So 4 + 2A = 18 which produces
A = 7.
3) The largest amount that can be made is $0.49. Using the given set of coins, any amount from $0.01 to $0.49 can be made. Therefore there are 49 different amounts that can be made.
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 email@example.com.
Walton High School
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 two-digit number AB as shown below. What is the value of C?
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?
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
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?
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.