1) A box contains over 100 marbles. The marbles can be divided into equal shares among 6, 7, or 8 children with 1 marble left over each time. What is the least number of marbles that the box can contain?
2) A fisherman sold some big fish at $4 each and twice as many small fish at $1 each. He received a total of $72 for the big and small fish. How many big fish did he sell? 3) In the addition example below, different letters represent different digits. What digit does A represent? A A + A A C A B Solutions: 1) Use a simpler problem. Instead of one marble being left over each time, assume that no marbles were left over each time. Then the least number of marbles is the LCM(6, 7, 8) which is equal to 3 x 7 x 8 = 168. However, since one marble should be left over each time, the least number of marbles in the box is 168 + 1 = 169. 2) For each big fish sold for $4, two small fish were sold for $1 each. One big fish and two small fish were sold for a total of $6. Since the total received for big and little fish was $72, there were 72/6 = 12 sets of 1 big fish and 2 little fish sold. Therefore 12 big fish were sold. 3) Method 1: A has to be 5 or more. Otherwise the sum will not be a threedigit number. In the second column, A + A + 1 ends in A. Then A is odd, so A = 5, 7, or 9. The only value that checks is 9. Therefore A is 9. Method 2: Use expanded notation: AA = 10A + A = 11A. Then AA + AA = 22A. Therefore C must be 1 and CAB = 100 + 10A + B. Then 22A = 100 + 10A + B. If we subtract 10A from both sides of the equality, the result is 12A = 100 + B. Since 12A has to be greater than 100, A has to be 9.
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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 + 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. 1) The numbers 2, 4, 6, and 8 are a set of four consecutive even numbers. Suppose the sum of five consecutive even numbers is 320. What is the smallest of the five numbers?
2) Twelve people purchased supplies for a tenday camping trip with the understanding that each of the twelve will get equal daily shares. They are then joined by three more people, but make no further purchases. How long will the supplies then last if the original daily share for each person is not changed? 3) Amy can mow 600 square yards of grass in 1 1/2 hours. At this rate, how many minutes would it take her to mow 600 square feet? Solutions 1) Method 1: The middle number of an odd number of consecutive numbers is always the average of the set. Then the average of the numbers is 320/5 or 64 which also is the third or middle number. Count back by twos. The required number is 60. Method 2: Represent the middle number by n. Then the five consecutive even numbers are n  4, n  2, n, n + 2, and n + 4. The sum of the five numbers is 5n. Since 5n = 320, n = 64. Thus n  4, the first number, is 60. 2) Since each person of the original group had 10 daily shares, the total supplies are equivalent to 120 daily shares. When 3 people join the group, the total number of people becomes 15. Then each person in the new group will have 120/15 or 8 daily shares. The supplies will last 8 days. 3) Method 1: 1 square yard = 9 square feet. Then 600 square yard = 600 x 9 square feet. Since 600 square feet is 1/9 of 600 x 9 square feet, the time needed to mow 600 square feet is 1/9 of the time required to mow 600 square yard. Therefore, 1/9 of 1 1/2 hours is 1/9 x 3/2 = 1/6 hour or 10 minutes. Method 2: Since 9 square feet = 1 square yard, 1 square foot = 1/9 square yard. Then the time needed to mow 1 square foot is 1/9 of the time needed to mow 1 square yard. Therefore, 600 square feet will require 1/9 of 1 1/2 hour or 1/6 hour or 10 minutes. 1) A bicyclist wants to make a 600mile trip on his twowheel bicycle. He has a spare wheel which is used to replace either of the other two wheels. Suppose each of the three wheels is to have the same mileage for the trip. How many miles should each wheel travel?
2) How many even numbers between 1 and 101 are multiples of 3? 3) In the subtraction problem below, all five of the digits 3, 5, 6, 7, and 9 are to be placed, one in each box. What is the smallest difference that can be the result? Solutions 1) No matter which two wheels are actually used at any time during the 600mile trip, the total miles traveled by those wheels together is 1200 miles. Since three wheels are to share that total equally, each will travel 400 miles. 2) The even multiples of 3 are 6, 12, 18, . . . The largest multiple of 6 less than 101 is 16 x 6 = 96. Multiples of 6 less than 101 are: 1 x 6, 2 x 6, 3 x 6, 4 x 6, . . . , 16 x 6. There are 16 even multiples of 3 between 1 and 101. 3) Make the 3digit number as small as possible and the 2digit number as large as possible, as shown at the right. The smallest possible difference is 259. 3 5 6  9 7 2 5 9 1) I have exactly ten coins whose total value is $1. If three of the coins are quarters, what are the remaining coins?
2) A group of 21 people went to the county fair with 9 people on a stagecoach and 3 people in a buggy. On the return trip, 4 people rode in each buggy. How many people returned on the stagecoach? 3) In the addition problem below, each letter stands for a digit and different letters stand for different digits. What digits do the letters H, E, and A each represent? H E H E H E + H E A H Solutions 1) We have to find 7 coins whose value is $0.25. If the coins were nickels, their total value would be too large. There must be at least 5 pennies. Then we need two coins whose value is $0.20. The coins are dimes. Therefore the remaining coins are 5 pennies and 2 dimes. 2) Going to the fair, 12 people rode in the buggies. Since 3 people rode in each buggy, there were 4 buggies. On the return trip, 4 people rode in each buggy. Then 16 people rode in the buggies. Since the total number of people was 21, 5 rode in the coach. 3) In the tens column, H is less than 3. Otherwise the sum would be a threedigit number. H = 1 or 2. In the units column, the sum of 4 Es is an even number. Then H in the sum must be 2. It follows that E must be either 3 or 8. If E = 8, the sum will be a threedigit number. Thus E = 3, H = 2, and A = 9. 1) In the following sequence of numbers, each number has one more 1 than the preceding number: 1, 11, 111, 1111, 11111, ... What is the tens digit of the sum of the first 30 numbers of the sequence?
2) When asked how many gold coins he had, the collector said: If I arrange them in stacks of five, none are left over. If I arrange them in stacks of six, none are left over. If I arrange them in stacks of seven, one is left over. What is the least number of coins he could have? 3) In the subtraction problem below, each letter represents a digit, and different letters represent different digits. What digit does C represent? A B A  C A A B Solutions 1) The ones column of the 30 numbers contains 30 ones making a sum of 30. Thus, the ones digit of the sum is 0, carry 3. The tens column contains 29 ones. Its sum is 29 plus the 3 from the "carry," making 32. Therefore the tens digit of the sum is 2. 2) The number of coins must be a multiple of 30. The multiples of 30 are 30, 60, 90, 120, 150, and so on. The smallest of these multiples that leaves a remainder of 1 when divided by 7 is 120. 3) It is clear that B = 0 and A = 1. Substitute those numbers for letters as shown below. Therefore C is 9. 1 0 1  C 1 1 0 to edit. 1) A girl bought a dog for $10, sold it for $15, bought it back for $20, and finally sold it for $25. Did the girl make or lose money, and how much did she make or lose?
2) The average of five weights is 13 grams. This set of five weights is then increased by another weight of 7 grams. What is the average of the six weights? 3) Each of the boxes in the figure is a square. How many different squares can be traced using the lines in the figure? Solutions 1) She paid out $10 + $20 = $30. She received $15 + $25 = $40. She made $10. 2) The average of the five weights is 13 grams. Then the total weight of the five weights is 5 x 13 or 65 grams. The sixth weight increases the total to 72 grams. The average of the six weights is 72/6 or 12 grams. 3) There are three different sizes for the squares that can be traced in the figure: 1 x 1, 2 x 2, and 3 x 3. The table shows how many squares can be traced for each size. 
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