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.
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AuthorAmanda Lovett Archives
December 2017
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