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Mathematical Olympiads For Elementary & Middle Schools November 12, 2019 Name: ____________________________________________________ 1A Add: 8883 + 8838 + 8388 + 3888. 1B Five students entered a contest where they guessed how many marbles were in a jar. The person with the closest guess won the contest. • Ashley guessed 98 marbles. • Beth guessed 105 marbles. • Candace guessed 109 marbles. • Dennis guessed 113 marbles. • Edward guessed 115 marbles. Two of the students were 3 away from the actual number of marbles. Who won the contest? 1C Each letter in the following addition problem represents a different digit and each of the same letter represents the same digit. No letter can be either 5 or 9. Determine the sum PLAN in this cryptarithm. not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is to be done on the pages provided. No additional scrap paper is to be used. Answers must be placed in the corresponding boxes in the answer column. 5 A 9 5 A 9 PLAN
Name: _________________________________________________________________ 5 13 can be written as the decimal 0.384615384615… where the digits 384615 keep repeating. What is the 2019th digit to the right of the decimal point? 1E Khadijah is thinking of a 5-digit number using all the digits 0, 1, 2, 3, and 4. The number is divisible by 5, by 8, and by 11. The leftmost digit of the 5-digit number is not 4. What is the number? Answer Column 1A 1B 1C 1D 1E Do Not Write in this Space. For PICO’s Use Only. SCORE: –Page may be folded along dotted line– –Page may be folded along dotted line.
Mathematical Olympiads For Elementary & Middle Schools December 10, 2019 Name: ____________________________________________________ 2A Evaluate: 98 – 87 + 76 – 65 + 54 – 43 + 32 – 21. 2B In the figure, how many squares (of any size) can be drawn along existing lines such that the number in a square or the sum of the numbers inside each larger square is even? 2C What is the value of 1974 6 + 1969 4 + 45 6 + 50 4? not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is to be done on the pages provided. No additional scrap paper is to be used. Answers must be placed in the corresponding boxes in the answer column. 1 2 4 5 7 8
Name: _________________________________________________________________ 2D At the corner diner, lunches have one set price and dinners have another set price. Five lunches and three dinners cost $70, and three lunches and five dinners cost $74. Determine the cost of one lunch. 2E In the following cryptarithm, each letter represents its own digit. Determine the greatest value of the 4-digit number OUCH. Answer Column 2A 2B 2C 2D $ 2E Do Not Write in this Space. For PICO’s Use Only. SCORE: –Page may be folded along dotted line– –Page may be folded along dotted line.= OUCH=
Mathematical Olympiads For Elementary & Middle Schools January 14, 2020 Name: ____________________________________________________ 3A Find the sum: 3B In the list of numbers 3, 8, 13, 18, …, 98, there are 20 numbers, and each number after the first is five more than the previous number. All of the numbers are added together to give a sum. What is the units digit of the sum? 3C In a family, there are four children. Adam’s age is the sum of Beth’s and Carol’s. Four years ago, David’s age was the sum of Beth’s age then and Carol’s age then. Eight years ago, Adam’s age was twice David’s age then. Who is the oldest child in this family? not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is to be done on the pages provided. No additional scrap paper is to be used. Answers must be placed in the corresponding boxes in the answer column. 202222 220222 222022 222202
Name: _________________________________________________________________ 3E It takes 12 workers exactly 2 hours to build a total of 60 widgets. Working at the same rate, how many widgets could 4 of those workers build in 8 hours? Answer Column 3A 3B 3C 3D 3E Do Not Write in this Space. For PICO’s Use Only. SCORE: –Page may be folded along dotted line– –Page may be folded along dotted line.
Mathematical Olympiads For Elementary & Middle Schools February 11, 2020 Name: ____________________________________________________ 4A Evaluate: 732 – 935 + 868 – 265. 4B A 3-digit counting number has each of the following properties: * The three digits are in increasing order. * The sum of the digits is 9. * The number is a multiple of 15. What is the number? 4C The numbers 10, 11, 12, …, 98, 99 are written as one big 180-digit number (1011121314 … 979899). All the even digits are then removed. How many digits remain? not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is to be done on the pages provided. No additional scrap paper is to be used. Answers must be placed in the corresponding boxes in the answer column.
Name: _________________________________________________________________ 4D What is the least 2-digit counting number that is a multiple of 3, that is one less than a multiple of 4, and that is one more than a multiple of 5? 4E What is the maximum number of 3 3 2 bricks that will fit in a box that has dimensions 13 8 4 without exceeding the dimensions of the box? Answer Column 4A 4B 4C 4D 4E Do Not Write in this Space. For PICO’s Use Only. SCORE: –Page may be folded along dotted line– –Page may be folded along dotted line.
Mathematical Olympiads For Elementary & Middle Schools March 10, 2020 Name: ____________________________________________________ 5A What is the value of 47 12 + 12 53 + 53 8 + 8 47? 5B Don, Jon, and Ron are each thinking of a 2-digit prime number. The numbers are different. Don’s number is the least and Ron’s number is the greatest. If each of their numbers has digits that total 8, what is Jon’s number? 5C Sally leaves her house and jogs along a straight road. After one hour she has gone 4 miles. Then, Sally slows to a brisk walk at a constant rate. After the first hour of walking in the same direction, Sally is 6.5 miles from her house. How many miles from her house will she be after two more hours of walking? not have a pen or pencil in your hand while the PICO reads the set of questions to the class. Calculators are not permitted. All work is to be done on the pages provided. No additional scrap paper is to be used. Answers must be placed in the corresponding boxes in the answer column.
Name: _________________________________________________________________ 5D A rectangular prism (box) has volume 240 cubic cm. A smaller rectangular prism is removed from the larger prism. The lengths of the edges of the smaller prism are one-half the lengths of the corresponding edges of the larger prism. How many cubic centimeters is the volume of the remaining solid? 5E There is an octahedral (8-faces) die whose faces each have one different number from the set {1, 2, 3, 4, 5, 6, 7, 8}. The die is rolled twice, and the number appearing on top after the first roll is multiplied by the number appearing on top after the second roll. What is the probability that this product is divisible by 9? Answer Column 5A 5B 5C 5D 5E Do Not Write in this Space. For PICO’s Use Only. SCORE: –Page may be folded along dotted line– –Page may be folded along dotted line.
Mathematical Olympiads For Elementary & Middle Schools November 12, 2019 The ones, tens, hundreds, and thousands place digits each consist of 3, 8, 8, and 8 in some order. Using 3 + 8 + 8 + 8 = 27 in each place value, the problem becomes 27 + 270 + 2700 + 27000, which equals 29997. In each place value there are three digits that are 8 and one that is a 3. Split each three into three 1’s and add one 1 to each 8. The problem is the same as 9990 +9909 + 9099 + 999 = 29997. FOLLOW-UP: Find the sum of 6423 + 4236 + 2364 + 3642. [16665] Since two of the students were 3 away from the actual number, find the two guesses that have a difference of 6. The two guesses belong to Candace and Edward, since 115 – 109 = 6, which means there were 112 marbles. The guess of 113 marbles is closest to 112, therefore Dennis won the contest. Add 3 and subtract 3 from each guess to find the two students who had the same guess when they were 3 away from the actual number. Student Guess +3 Ashley 98 101 Beth 105 108 Candace 112 Dennis 113 116 Edward 115 The number 112 is 3 more than Candace’s guess and 3 less than Edward’s guess. There are 112 marbles in the jar, so Dennis is the winner with his guess of 113. The ones column adds to 27. The tens digits in the three addends is the same as the tens digit in the sum so 3A + 2 equals A + 10 or A + 20. It follows that 2A equals 8 or 18 so A equals 4 or 9. Since A ≠ 9, A = 4 and the addends are each 549. The sum is 3 549 = 1647. SOLUTIONS AND ANSWERS1A 29997 1B Dennis 1C 1647 1D 4 1E 31240
Examine the product of 3 times each addend 3 509 = 1527, 3 519 = 1557, 3 529 = 1587, 3 539 = 1617, 3 549 = 1647, etc. Reject the first four possibilities since the tens digit in the product does not equal the tens digit in the addends. The number 1647 satisfies all the conditions. FOLLOW-UP: In the following cryptarithm, each letter represents a different digit from 1 through 9. If T is 5, find any possible value for PETS. [1357 and 1258 are two solutions.] Since every six digits of the decimal repeats, divide 2019 by 6 to get 336 with a remainder of 3. Use the remainder to find the 2019th digit of the decimal. The third digit to the right of the decimal is 4, therefore the 2019th digit will be 4. Since 384615 repeats, the digit 5 will repeat in every 6th position. Multiples of 6 are divisible by 2 and 3. Numbers divisible by 3 have the sum of their digits as a multiple of 3. Numbers divisible by 6 have that property but must also be even. The number 2019 is divisible by 3 since 2 + 0 + 1 + 9 = 12 but 2019 is not even. If we subtract 3 from 2019, we get 2016 which is even and still divisible by 3. Since we want the 2019th digit, it will be 3 digits beyond the digit 5 found in the 2016th position. That digit is 4. FOLLOW-UP: The fraction 1/54 is written as a decimal number. What is the 2020th digit to the right of the decimal point? [5] If a number is divisible by 5, then the units digit must be a 0 or 5. Since 5 is not one of the usable digits, the 5-digit number must end in a 0. If a number is divisible by 11, then the alternating sum and difference of the digits from left to right must also be divisible by 11 (Example: 95381 is divisible by 11 since 9 – 5 + 3 – 8 + 1 = 0 which is divisible by 11.). Using the digits 1, 2, 3, and 4, the only possible alternating sum that would be divisible by 11 is 0. To get an alternating sum of 0, the digits 1 and 4 should alternate with 2 and 3. This could result in 12430, 21340, and 31240 (42130 cannot be a possible solution because the first digit is 4). To check which of these is divisible by 8, see if the last three digits of the number are divisible by 8. Since 240 is divisible by 8, the 5-digit number is 31240. Make a list of possible 5-digit numbers that fit the parameters of the problem. The units digit must be 0 in order for the number to be divisible by 5 and even. Consider the expansion of the 5-digit number abcde: abcde = 10000a + 1000b + 100c + 10d + e. Since the place values 10000 and 1000 are each divisible by 8 the number 10000a + 1000b is divisible by 8. Determine which 3-digit numbers using the given digits are divisible by 8: 120, 240, and 320. For each of these there are at most two possible choices for the digits a and b. Test to see which of the numbers is also divisible by 11: 34120, 13240, 31240, and 14320. The only one divisible by 11 is 31240. FOLLOW-UP: Find the value of the digit A in the number 3020202A that would make the number divisible by 9 and by 11. [9] D O G P E T S
Mathematical Olympiads For Elementary & Middle Schools December 10, 2019 Copyright © 2019 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. (98 – 87) + (76 – 65) + (54 – 43) + (32 – 21) = 11 + 11 + 11 + 11 = 4 × 11 = 44. Plusses: 98 + 76 + 54 + 32 = (90 + 70 + 50 + 30) + (8 + 6 + 4 + 2) = 240 + 20 = 260 Minuses: 87 + 65 + 43 + 21 = (80 + 60 + 40 + 20) + (7 + 5 + 3 + 1) = 216 Subtract: 260 – 216 = 44 10 ((9 – 8) + (7 – 6) + (5 – 4) + (3 – 2)) = 40 (8 – 7) + (6 – 5) + (4 – 3) + (2 – 1) = 4 Therefore 98 – 87 + 76 – 65 + 54 – 43 + 32 – 21 = 44. FOLLOW-UP: Evaluate: 91 – 82 + 73 – 64 + 55 – 46 + 37 – 28. [36] For a sum to be even, there needs to be an even number of odd numbers (and it does not matter how many even numbers there are). The only 1 × 1 squares are 2, 4, 6, and 8. Then, all of the 2 × 2 squares have two odds apiece, on a diagonal. Note: the 3 × 3 square has five odd numbers and would not have an even number sum. So, there is a total of 4 + 4 + 0 = 8 squares. Draw the squares that meet the criteria. FOLLOW-UPS: (1) For the same figure, how many rectangles (of any size) can be drawn along existing lines such that the number or the number sum inside the rectangle is even? [12] or odd? [24] (2) In the figure, how many squares (of any size) can be drawn along existing lines, such that the number or the number sum inside the square is divisible by 3? [6] (1974 × 6) + (45 × 6) + (1969 × 4) + (50 × 4) = (1974 + 45) × 6 + (1969 + 50) × 4 = 2019 × 6 + 2019 × 4 = 2019 × (6 + 4) = 2019 × 10 = 20190. SOLUTIONS AND ANSWERS2A 44 2B 8 2C 20190 2D $8 2E 954312452356568945784628
Lining up the products (as areas), we almost get a 4038 × 6 rectangle. The area of the shaded rectangle (2019 × 2 = 4038) needs to be removed. In total, (4038 × 6) – (4038 × 1) = 4038 × (6 – 1) = 4038 × 5 = (2 × 2019) × (½ × 10) = (2 × ½) × (2019 × 10) = 1 × 20190 = 20190. FOLLOW-UPS: (1) Evaluate 576 × 17 + 582 × 15 + 1424 × 17 + 1418 × 15. [64,000] (2) Evaluate (1976/6) + (1974/3) + (1969/2) + (43/6) + (45/3) + (50/2). [2019] If turning two lunches into dinners raises the cost $4 then a dinner must cost $2 more than a lunch. Then turning three dinners into lunches should reduce the cost by $6. It follows that eight lunches should cost $64. Therefore, one lunch should cost $8. In total, there are (5 + 3) = 8 lunches, (3 + 5) = 8 dinners, and together the cost is ($70 + $74) = $144. Divide this by 8 to get that 1 lunch + 1 dinner = $18. So, 3 lunches + 3 dinners = (3 × $18) = $54. Compare this with the cost of 5 lunches + 3 dinners = $70 to see that the extra (5 – 3) = 2 lunches cost ($70 – $54) = $16, or that one lunch costs $8. FOLLOW-UP: At the down-the-block diner, lunches have one set price and dinners have another set price. If four lunches and three dinners cost $58, and five lunches and four dinners cost $75, determine the cost of one lunch. [$7] To maximize OUCH, let O = 9, so S = 8. U cannot be 7 or 6 because it would make T = 8 = S. Let U = 5, so T = 7. In the tens place, U + O = C becomes 5 + 9 = 14, so C = 4. Of the remaining digits: 0, 1, 2, 3, and 6, only H = 3 (from 1 + 2) works. Thus, OUCH = 9543. First try: OUCH = 9876 which results in S = U = 8. (In fact, if O = 9, then S = 8.) Second try: OUCH = 9765 with S = 8. But the only way to get U = 7 is if T = 8 (which is already S) or T = 3 (which fails to cause regrouping in the thousands place). Third try: OUCH = 9675 with S = 8. Similarly, there is no good value for T. Fourth try: OUCH = 9576 with S = 8. Here, T = 7 would work if C wasn’t 7. So, adjust to try OUCH = 9564 with S = 8 and T = 7. Here, 875B + 79E = 9564. That doesn’t work. Adjust again to try OUCH = 9546 with S = 8 and T = 7. Here, 875B + 79E = 9546. There are no possible pairs of B and E that work. Adjust again to try OUCH = 9543 with S = 8 and T = 7. Here, 875B + 79E = 9543, which works because B and E can be 1 and 2, in some order. FOLLOW-UP: Knowing that 0 can never be a starting digit of a number, find the least value of OUCH in the given cryptarithm. [2034]
Mathematical Olympiads For Elementary & Middle Schools January 14, 2020 Copyright © 2019 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. There are four 2’s in the ones, tens, hundreds, thousands, and ten-thousands columns. The hundred-thousands column contains five 2’s. The sum of each of the ones, tens, hundreds, thousands, and ten-thousands column is 8. The hundred- thousands column sums to 10. The resulting sum is 1088888. FOLLOW-UP: Find the sum 505555 + 550555 + 555055 + 555505 + 555550. [2722220] There are 20 numbers in the list and each subsequent number increases by 5, resulting in a pattern of 3, 8, 3, 8, 3, 8 … in the units place. Add 3 + 8 = 11. Since there are 10 equal groups of 11, the units digit will be the units digit of the product 10 11 = 110, which equals 0. A quick way to add all the numbers in the list is to write the numbers twice and add: 101 + 101 + 101 + 101+ … + 101 + 101 + 101 + 101 The total will equal 20 101 = 2020. This is twice the total we need or 1010. Therefore, the units digit is 0. FOLLOW-UPS: (1) Find the units digit of the sum of all the positive multiples of 4 that are less than 100. [0] (2) Find the tens digit for the sum in part (1). [0] Adam’s age = Beth’s + Carol’s ages David’s age = Beth’s + Carol’s ages (4 years ago) Adam’s age = 2 David’s age (8 years ago) Therefore, Beth and Carol are younger than both Adam and David. Since Adam’s age was twice David’s age 8 years ago, Adam must be the oldest child. SOLUTIONS AND ANSWERS3A 1088888 3B 0 3C Adam 3D 18 3E 80 202222 220222 222022 222202 1088888
Could David be the oldest child? No, because Adam was twice David’s age 4 years ago. Could Carol be the oldest child? No, since Adam’s age is the sum of Carol’s and Beth’s ages. Could Beth be the oldest child? No, due to the reasoning above. Thus, Adam must be the oldest child. FOLLOW-UP: If Beth and Carol are twins, what are their current ages? [8] The number of 1 1 squares is 12. The number of 2 2 squares is 1 (the inner square). The number of 3 3 squares is 4. The number of 4 4 squares is 1 (the outermost square). The total number of squares is 12 + 1 + 4 + 1 = 18. Count the number of squares that contain the upper left corner 1 1 square: there is the square itself, a 3 3, and a 4 4 square, for a total of 3 squares. Each corner square has the same property, but we do not wish to count the 4 4 square more than once. Therefore, there are 3 + 2 + 2 + 2 = 9 squares that contain corner squares. Count the number of squares that contain an edge square but no corner square: there are the 8 squares each measuring 1 1. Finally, the last square to count is one that does not contain either an edge or a corner square: there is only the one 2 2 square in the middle. Thus, we have a total of 9 + 8 + 1 = 18 squares. FOLLOW-UP: Find the total number of rectangles in the diagram. [52] Find the rate for each worker, 60 widgets/12 workers = 5 widgets/worker in 2 hours. This means that each worker builds 5 widgets in 2 hours or 20 widgets in 8 hours. Therefore, 4 workers can build 4 20 = 80 widgets in 8 hours. Make problem simpler by dividing by 3 to create a chart for 4 workers only. Number of Workers Cumulative Hours 4 2 4 4 4 6 4 8 FOLLOW-UP: Find the number of minutes it would take 1 worker to build 1 widget. [24 minutes]
Mathematical Olympiads For Elementary & Middle Schools February 11, 2020 Copyright © 2019 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. two numbers being subtracted. 732 + 868 = 1600 935 + 265 = 1200 Subtract the second result from the first: 1600 – 1200 = 400. (732 + 868) – 935 – 265 = (1600 – 935) – 265 = 665 – 265 = 400. FOLLOW-UP: Mr. Bear had $157 in the bank. He borrowed enough money to buy something for $249. That week he earned $529. He paid off his debt and put the rest in the bank. How much does he have in the bank? [$437] 7KHQXPEHULVDPXOWLSOHRIVRWKHXQLWVGLJLWPXVWEHRU6LQFHWKHGLJLWV DSSHDULQLQFUHDVLQJRUGHUWKHXQLWVGLJLWFDQQRWEHVRLWPXVWEHILYH6LQFHWKH WKUHHGLJLWVVXPWR DQGWKHXQLWVGLJLWLVDOUHDG\WKHKXQGUHGVDQGWHQVGLJLWV PXVWVXPWR7KHGLJLWVPXVWEHGLIIHUHQWLQRUGHUWREHLQLQFUHDVLQJRUGHUVRWKH WZRGLJLWVPXVWEHDQG7KHUHIRUHWKHVROXWLRQPXVWEH135. The only 3-digit numbers whose digits add up to 9 and that are in increasing order are: 126, 234, and 135. The only one that is a multiple of 15 is 135. properties: (a) Its four digits are in decreasing order.=(b) The sum of the digits is= 18. And (c) The number is a multiple of 15.=[9810] 7KHUHDUH VHWVRIWZRGLJLWQXPEHUVIURPWR 7KLVPDNHVGLJLWV 7KHILUVWVHWRIGLJLWQXPEHUVEHJLQQLQJZLWKDQRGGQXPEHU ³RGGEHJLQQHUV´ LVVHWVRIWZRGLJLW³RGGEHJLQQHUV´PDNLQJRGGGLJLWV 7KHILUVWVHWRIGLJLWQXPEHUVEHJLQQLQJZLWKDQHYHQQXPEHU ³HYHQEHJLQQHUV´ LV1, 22, 23, 24, 25, 26, 27, 28, 29. Count odd digits only. There are 5. There are 4 sets of two-digit “even beginners,” making 20 odd digits. Thus, there are 75 + 20 = 95 odd digits of the 180 digits. FOLLOW-UP: The numbers 100, 101, 102, …, 198, 199 are written as one big number. All the odd digits are then removed. How many digits remain? [100] SOLUTIONS AND ANSWERS4A 400 4B 135 4C 95 4D 51 4E 20
7KHUHDUHIHZHUPXOWLSOHVRIWKDQRIRU6REHJLQZLWKWKHFOXHUHODWHGWRWKHPXOWLSOHVRI 0XOWLSOHVRIKDYHRULQWKHXQLWVSODFH1XPEHUVWKDWDUHRQHPRUHWKDQDPXOWLSOHRIQHHGWRHQG LQDRUD0XOWLSOHVRIDUHHYHQQXPEHUV2QHOHVVWKDQDPXOWLSOHRIZRXOGEHDQRGGQXPEHU 7KXVWKHQXPEHUPXVWKDYHLQWKHXQLWVSODFH0XOWLSOHVRIKDYHGLJLWVWKDWDGGWRRUDPXOWLSOHRI 6LQFHZHZDQWWKHOHDVWQXPEHUWKDWKDVWKHVHDWWULEXWHVOLVWWKHWZRGLJLWQXPEHUVIURPOHDVWWRJUHDWHVW WKDWKDYHDLQWKHXQLWVSODFHZKRVHGLJLWVDGGWRDPXOWLSOHRI 2IWKHVHDOODUHPXOWLSOHVRI DQGDOODUHRQHPRUHWKDQDPXOWLSOHRI7KHRQO\RQHWKDWLVRQHOHVVWKDQDPXOWLSOHRILV51. Cross out those that are not a multiple of 3: 11, 31, 51, 71, 91 → The only number not eliminated is 51. FOLLOW-UP: How many 3-digit numbers are there that are one more than a multiple of 9 and 1 less than a multiple of 10? [10] $ERWWRPOD\HURI 3 2 (height) gives 4 bricks along the long side and 2 bricks along the width. This leaves a unit of 1 on the length and 2 on the width. We now have 8 bricks on the bottom layer. We have room for another layer in height, for 16 bricks. Finally, along the width we have a 2 3 3. This will fit another 4 bricks. In total 20 bricks can fit in the box. Since we can fit 2 layers of bricks with the 3 3 side face down and a single layer of bricks with the 2 3 side face down, we can fit a total of 2(8) + 4 = 20 bricks in the 13 8 4 box. FOLLOW-UP: Using the 24 bricks with dimensions 3 3 2, compute the difference between the greatest area that can be covered by the bricks and the least area that can be covered by the bricks. [72 sq. units] Visualize the area of the bottom of the box.3ODFHDVPDQ\EULFNVDFURVV WKHERWWRPZLWKWKH 3 side face down. (8 bricks))LOOLQWKHHPSW\VSDFHZLWK EULFNVXVLQJWKH
Mathematical Olympiads For Elementary & Middle Schools March 10, 2020 Copyright © 2019 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. Regroup: 12 (47 53) 8 (47 53) (12 100) (8 100) 1200 800 2000 47 12 + 12 53 + 53 8 + 8 47 564 636 424 376 2000 FOLLOW-UPS: (1) What is the value of 47 12 12 53 – 53 8 – 8 47? [400] (2) What is the value of 33 12 12 66 – 66 8 – 8 33? [396] an ordered sub list of the primes. The 2-digit whole numbers with digit sum equal to 8: 17, 26, 35, 44, 53, 62, 71, 80. The primes from this list are: 17, 53, 71. From sentence #2: Don’s # Jon’s # Ron’s #, therefore Jon’s number is 53. Since all the numbers are prime, eliminate any even two-digit numbers. List the odd two-digit numbers that add up to 8: 17, 35, 53, 71. Thirty-five is not prime (multiple of 5) so the numbers are 17, 53 and 71. Ron’s is the greatest (71), Don’s number is the least (17) and Jon’s number must be 53. FOLLOW-UPS: Given the original problem, if the digit sum is 10, what is Jon’s number? [37] Sally jogs 4 miles in 1 hour. Walking, Sally covers an additional 2.5 miles in the next hour. Therefore, Sally’s walking rate is 2.5 mph. After 2 more hours of walking, Sally covers an additional 2 hours 2.5 mph 5 miles. Altogether, Sally has covered 4 2.5 + 5 11.5 miles in the 4 hours. Distance (miles) 4 2.5 2.5 Time (hours) 1 1 1 SOLUTIONS AND ANSWERS5A 2000 5B 53 5C 11.5 5D 210 5E 1/16Sally covered 4 + 2.5 + 2.5 + 2.5 = 11.5 miles in the 4 hours.
1 2L,1 2W,and1 2Hcubic cm. The remaining volume is 240 30 210 cm3 . Even-numbered dimensions for a 240 cubic cm rectangular prism could be length = 12, width = 10, and height = 2. (Volume = length width height.) Halving each of those dimensions, we get a volume of 6 5 1 = 30 cubic cm. Subtracting the removed volume, 240 – 30 = 210 cm3 . (Similarly, we could have chosen 4, 6, and 10; or 2, 4, and 30; or 2, 2, and 60; or 2, 6, and 20 for the original dimensions). Observe that the smaller prism is 1/8 of the entire prism. Therefore, the remaining volume will be 7/8 of the entire volume and (7/8) 240 = 210 cm3 . FOLLOW-UPS: Use the original problem and let each of the dimensions of the cutout be a positive integer: (1) What is the least possible surface area of the cutout rectangular prism? [62 square cm] (2) What is the greatest possible surface area of the cutout rectangular prism? [122 square cm] 1 16. 1 8. FOLLOW-UPS: (1) Beginning with the same octahedral die, what is the probability of rolling the die twice and having the product of the two values be a perfect square? [3/16] (2) Using the original problem, what is the probability that the product is not divisible by 9? [15/16] 363636