Solutions to Exercises on Page 86

Solutions to Exercises on Page 86 #. A number is a multiple of, 4, 5 and 6 if and only if it is a multiple of the greatest common multiple of, 4, 5 and 6. The greatest common multiple of, 4, 5 and 6 is 45 60. Between 000 and 9000 there are 8999-000=7999 integers (excluding 000 and 9000). 7999/60= with a fractional part. From 00 to 8999 there are multiples of 60 because both 00 and 8999 are not multiples of 60 and multiples of 60 are evenly distributed in the sequence with a distance of 60. Answer: #. From 8 am to 5 pm there are 9 hours. There are 6 5-minutes intervals. Because both the two endpoints count, 7 ferries left yesterday. Answer: 7. #. 50+90=40, 5+89=40,, 69+7=40. The numbers from 50 to 90 other than 70 can be paired so that only one of each pair can be included in the set. There are 0 such pairs. 70, 9, 9,, 00 can be freely included in the set without violating the condition. The maximum such set will have 0++0= elements. Answer:. #4. 00+50=0+49= =4+6=450. The numbers from 00 to 50 other than 5 can be paired so that only one of each pair can be included in the set. There are 5 such pairs. 5, 5, 5,, 00 can be freely included in the set without violating the condition. The maximum such set will have 5++50=76 elements. Answer: 76. #5. 00 5. The number of factors is 9. Answer: 9 #6. 80 5. Even factors of 80 must be in the form a b c 5 with a = or, b =0, or, c=0 or. The number of even factors is. We can also count the number of odd factors first. The odd factors must be factors of 5. Answer: #7. We will understand that divisors are the same as factors. We will just count the odd part of 0!. From to 0, there are multiples of ; there are multiple of. There are multiples of 5 and there is multiple of 7. So the odd part of 4 4 the factorization of 0! is 5 7. The number of factors of 5 7 is 5 0. Answer: 0 #8. There are 40 multiples of 5 in the first 0 positive integers and half of 04,Huasong Yin

them are odd. So there are 0 multiples of 5 in the sequence,, 5,, 0. 0/5=80+. There are 80 multiples of 5 in the first 0 positive integers and half of them are odd. So there are 40 multiples of 5 in the sequence,, 5,, 0. 0/5=6+. There are 6 multiples of 5 in the first 0 positive integers and half of them are odd. So there are 8 multiples of 5 in the sequence,, 5,, 0. 0/65=+. There are multiples of 4 5 in the first 0 positive integers and only one of them is even. So there are multiples of 4 5 in the sequence,, 5,, 0. n 0 40 8 5. Answer: 5. #9. There are multiples of from, to 99. 6 of them are even. So there are 7 multiples of in the sequence,, 5,, 99. There are multiples of but 5 of them are even. So there are 6 multiples of in,, 5,, 99. There are multiples in,, 5,, 99. There are multiples of 4 in,, 5,, 99. n 7 6 6. Answer: 6 #0. Use the same method as above, but now we could have formulated a formula for the maximum prime power n p in m!. n m m m p p p 04,Huasong Yin, where x is the integer floor function. 04 04 04 n 87 4 5 7 7 7 Answer: #. It is easier to count the numbers that do not contain digit. The first digit has choices; the second digit has 9 choices: the third digit has 9 choices also. By the product rule for counting, there are 9 6 numbers from 00 to 99

that do not contain digit. Excluding 00, there are 6. There are 99 total from 0 to 99. 99-6=8. Answer: 8 #. In the row, we cannot have three consecutive seats empty. In each end, we cannot have two consecutive seats empty. To optimize, we assume that each end has seat empty. From left to right, counting from the second seat, assume that the seating is one occupied followed by two empty seats. 96/=. We can arrange occupied seats so that the next person will seat to someone. (This is equivalent to put the 96 seats in a circle!). Answer:. #. If the three binomials are mx n y, mx n y and mx ny, then m m m 5 and n n n 7. Assume that m m m. The solutions for m, m, ) are (,,), (,,). ( m The solutions for n, n, ) are (,,5), (,,4), (,,), (,,) if the three ( n number are in order. We will count by case: ( m m, m, ) n, n, ) ( n Explanation: m m m, and permutate n, n, ) ( n Count (,,) (,,5) 5 can be any position (,,4)! permutations 6 (,,) (,,) (,,) 4 similar cases as above. similar The total number of count is 6 0. Answer: 0 #4. To maximize the number of intersection points, we can arrange the triangle to intersect a circle in 6 points. Move the circle slightly we get a distinct congruent circle that will intersect the triangle also in 6 points and it will intersect the original circle in two points. Total is 4. Two distinct circles cannot intersect at more than points. So 4 is the maximum number of intersection points. Answer: 4. #5. Prime digits are,, 5 and 7. An integer is divisible by 9 if the sum of digits is divisible by 9. 04,Huasong Yin

If the integer has only two digits, there are only two such integers: 7 and 7. If the integer has three digits, the sum of the digits can only be 9 or 8. If the three digits are the same, the integer is. If the three digits are different, there is no solution because ++5+7=7, 7-8=, 7-9=8 and both and 8 are not prime digits. If only two digits are the same, we find (,,5) only. This will give integers. So there are 4 solutions for -digit integers. If the integer has four digits, the sum of the digits can only be 9, 8 or 7. The four digits cannot be the same and they cannot be distinct. If three of them are the same, then the sum of the digits is a+b, b must be and a can be and 5. We get 8 solutions. If only two digits are the same, we get (,,7,7). This will give us C 6 solutions. Total in the four-digit case is 8+6=4. 4 +4+4=0. Answer: 0. #6. Case : one digit. There are such numbers: and 9. Case : two digits. If the first digit a=, 5, 9, a, a, a5, a7, a9 have exactly multiples of. If the first digit a=, 7, a, a, a5, a7, a9 have exactly multiple of. The total count in this case is 6+=8. Case : three digits. (a): If the first digit a= or 9, there are 8 6 solutions by case. (b): If the first digit is or 7, the sum of the last two digits must be b+c=, 8, 4 (sum is even). The solutions for (b, c) are (,), (7,7), (,7), (,5), (5,9) in order. Consider different ordering, there are 8 ordered pairs for (b, c). 8=6. (c): If the first digit is 5, the sum of the last two digits must be b+c=4, 0, 6. The solutions are (5,5), (,), (,9), (,7), (7,9) in order. There are 9 ordered pairs for (b,c). There are 6+6+9=4 solutions in the three-digit case. +8+4=5. Answer: 5 Solution (by Kevin Jiang) DIGITS ALL SAME:,,555,777,999 (5) SAME: 7(), 77(), 9(), 99() () NO SAME:5(6), 59(6), 57(6), 579(6) (4) DIGITS SAME:, 99 () NO SAME: 5(), 9(), 57() (6) DIGITS SAME:, 9 () 5++4++6+=5 04,Huasong Yin

#7. There are 6 consonants in the word RECTANGLE and they are all different. We understand that no letter can be used twice. The first letter of the word has 6 choices. Once the first one is chosen, the second letter has 5 choices. The third letter has 4 choices and the last letter has choices. The total number words is 6 54 60. Answer: 60 #8. The consonants in the word PERMUTATION are P, R, M, T, T and N with two T s. If T is not used twice, we have 5 4 0 four letter words. If T is used twice, we have C 6 ways to place the two T s and the two other 4 letters have 4 choices. 6 7. 0+7=9. Answer: 9 #9. The units digit can only be,, 5, 7, and 9. There are 5 choices for the units digit. If the units digits is, the thousands digit can only be 4. The hundreds digit has 8 choices and the tens digit has 7 choices after the tens digit is chosen. There are 7 8 56 numbers in this case. If the units digit is not, it has 4 choices. The thousands digit has choices. The hundreds digit has 8 choices and the tens digit has 7 choices after the tens digit is chosen. There are 4 78 448 numbers in this case. Total is 56+448=504. Answer: 504 #0. If a and b are two positive integers and a b, then a b ab b ab a a. So the smaller of the two numbers must be. There are 0 choices for the other number. As ordered pairs, there are 9 9 9 choices. The probability is. Answer: 00 00 #. To maximize the number of intersection points, any two lines must intersect and no three lines will intersect at the same point. Let a n be the maximum number of intersection points of n lines, then we have a a n n n. a, an ( n ) an ( n ) ( n ) an ( n ) ( n ) ( n ) n 89. a 9 6. Answer: 6 04,Huasong Yin

#. Any three points that are not collinear determine a triangle. There are 65 4 C 6 0 triangles using the six vertices of a hexagon. Answer: 0.! #. Exactly one child will receive two gifts. There are 5 ways to choose this child. There are C 5 ways to choose the two gifts. The remaining 4 gifts are given 6 to the 4 remaining kids. There are 4! ways to distribute the remaining 4 gifts. The total number of ways is 5 54! 554 800. Answer: 800 #4. If exact one child will receive more than one gift, this child will receive 654 gifts. Similar to above counting, we have 4 C 6! 4 6 0 4 480! ways to distribute the gifts. If two children will receive two gifts each, there are 6 ways to choose the two children. There are C 5 ways to choose the two gifts for the first child; there 6 are C 6 ways to choose the two gifts for the second child. There are two 4 ways to distribute the remaining two gifts. 6 56 080. 480+080=560. Answer: 560 #6. The four numbers must be different because C 6 and there are 6 different sums. Assume that a b c d. We have a b 5, a c 7, c d 4 and b d. c b. Choices for (a, b): (, 4), (, ). Choices for (c, d): (5, 9), (6, 8) For c-b= there are only two cases. Case : (a, b) = (, 4) and (c, d) = (6,8). We check that the six different sums are exactly 5, 7, 8, 0, and 4. Case : (a, b) = (, ) and (c, d) = (5, 9). The six different sums are 5, 7, 8,, and 4. Case is not our solution. (a, b)=(, 4), (c, d)=(6, 8). The product of the four numbers is 468 9. Answer: 9 #7. If the three digits are different, they can be chosen from {,,, 7}. There are 4 4 such positive integers. If the three digits are the same, 777 is the only number. The count is. If exactly two digits are the same, there are choices for the repeating digit and choices for the lonely digit and spots to place the lonely digit. 8. 4++8=4. Answer: 4 4 04,Huasong Yin

#8. If the three digits are different, they can be chosen from {,,, 4, 9}. There are 5 4 60 such positive integers. If the three digits are the same, 999 is the only number. The count is. If exactly two digits are the same, there are choices for the repeating digit and 4 choices for the lonely digit and spots to place the lonely digit. 4 6. 60++6=97. Answer: 97 #9. As in #6, assume that a b c d. We have a b 5, a c 7, c d 4 and b d. c b. Choices for (a, b): (, 4), (, ). Choices for (c, d): (5, 9), (6, 8) For c-b= there are only two cases. Case : (a, b) = (, 4) and (c, d) = (6,8). We check that the six different sums are 5, 7, 8, 0, and 4. Case : (a, b) = (, ) and (c, d) = (5, 9). The six different sums are 5, 7, 8,, and 4. Case is our solution. (a, b)=(, ), (c, d)=(5, 9). The product of the four numbers is 59 70. Answer: 70 #0. Assume that a b c d. We have a b 5, a c 8, c d 5 and b d. c b. Choices for (a, b): (, 5), (, ). Choices for (c, d): (6, 9), (7, 8) Only (a, b)=(, ) and (c, d)=(6, 9) have the property c-b=. The product of the four numbers is 69 4. Answer: 4 #. The product of three different numbers from {,,, 4, 5} will be even unless the three numbers are, and 5. Another choice of three different numbers will be, and 4 in order to have relative prime products. There are C C 0 ways to choose three different numbers from {,,, 4, 5 5 5}. The probability is 0 50. Answer: 50 #. If 6 is not chosen by any one, by the above count, there are two ways to have the two products relatively prime. If 6 is chosen by anyone, because, and 4 have common factors greater than with 6. Only two numbers and 5 are left. It is impossible to choose three numbers so that the product is relatively prime to Robert s product. The probability is C 6 C 6 0 0 00. Answer: 00 04,Huasong Yin

#. The first player comes out and chooses two players as group members. There are C 8 choices. 6 players remain. One player comes out and chooses two as the group members. There are C 5 choices. The total number of ways to 87 54 divide the team into -payer-groups is C 8 C5 80. Answer: 80 #4. The consecutive digits can be 0, 4, 45,, 6789. Each will give 4! 4 numbers. So there are 4 7 67 numbers consisting of 4 consecutive digits from to 9999. Let s count such numbers from to 0. If the four digits (allow 0 as the starting digit) are from 0, there are two cases. Case : The thousands digit is 0 or, there are! numbers. Case : the thousands digit is : The hundreds digit can only be 0; the tens digit can only be. 0 is the only number. +=. If the four digits are from 4, the thousands digit can only be. There are 6 such numbers. There are +6=9 such numbers from to 0. 67-9=48. Answer: 48 #5. The thousands digit can be any digit from to 9, and there are 9 choices for the thousands digit. The hundreds digit will be different from the thousands digit but it can be 0. There are also 9 choices for the hundreds digit. There are 8 choices for the tens digit. There are 7 choices for the units digit. There are 9 987 numbers from 000 to 9999 consisting 4 different digits. There are 9000 numbers from 000 to 9999 inclusive. The percentage is 9987 987 504 50.4% 9000 000 000 Answer: 50.4% #6. For any pizza topping, you can choose yes or no. There are 7 8 possible combinations of toppings for your pizza order. This is the number of subsets of a set of 7 elements. #7. If the four digits are in decreasing order, we just need to choose 4 different 4 0987 0987 digits. C 0 0. 4! 4 If the four digits are in increasing order, the digits cannot be 0. 4 9876 9876 C 9 6. 4! 4 0+6=6. Answer: 6 04,Huasong Yin

#8. and 9 are multiples of. Any different choice other than {, 5, 7} will contain a multiple of. There is only one pair of relatively prime products: 57 and 9. Answer: #9. This is similar to #8. Answer: #40. 4, 0, 6 are even. The only way to choose 4 numbers so that the product is odd is {, 7,, 9}. Any other choice of four different numbers to have a product relatively prime to 79 must be {, 4, 0, 6}. Answer: #4. See #. Answer: #4. Only three numbers in the set {, 4, 7, 0,, 6} are odd. {, 7, } is the only choice of three numbers to have an odd product. Another subset of three numbers must be chosen from, 4, 0 and 6 in order to have the product of them to be relative prime to 7. There are 4 ways to choose. Answer 4 #4. Any positive integer can be written as m n with m square-free. Let a m n, b m n with m and m square-free. a b is a square if and only if m m, i.e. they have the same square-free part. Since a, b are different, they will have different square part. We will just list all the subsets of {,,, 4, 5, 6, 7, 8, 9} with the same square-free part: {,, }, {, }, { }, { 5}, { 6}, { 7}. { a, b} can only be subsets of {,, } or {, }. C C 4. Answer: 4 #44. Similar to above, we will just list all the subsets with more than one element of {,,, 4, 5, 6, 7, 8, 9, 0,,,, 4, 5} with the same square-free part: {,, }, {, }, {, }. C C C 5. Answer: 5 #45. Here is the list all the subsets with more than one element of {,,, 4, 5, 6, 7, 8, 9, 0,,,, 4, 5, 6, 7, 8, 9, 0} with the same square-free part: {,,, 04,Huasong Yin 4 }, {,, }, {, }, {5, C C C C 6. Answer: 4 5 }.

#46. Here is the list all the subsets with more than one element of {,,, 4,, 40} with the same square-free part: {, 4 }, {,, }, {5, 5 }, {6,,, 4, 6 }, {7, 5, 6 }, {, 4 }, {0,, 4 }. C C C C C C C 5 6 4 8. Answer: 8 6 4, #47. The minimum sum is -+(-)+=-. The maximum sum is +4+5=. It's not hard to see all the numbers between can be the sum of three different numbers from the set. From - to there are 5 integers. Answer: 5 #48. A bus leaving at 0:00am from Memphis to Nashville will arrive at Nashville at :00pm. It will meet all the buses from Nashville leaving before :00pm. There are three buses leaving Nashville in each hour. There are 5 hours between 8:00am and :00pm. 5 5. Answer: 5 #49. This question is trivial because any arrangement will have at least two girls sitting next to each other. So we will change the question to at least two boys are next to each other. Let s count the opposite, i.e. no boys sitting next to each other. There are 6! ways to arrange the six girls from left to right. There are 7 spots(5 between, left and right) to insert the boys. So there are 7 654 ways to insert the boys. 6!7 654 70840 604800. 0! 604800 68800-604800 0400. Answer: 0400 #50. This question is also trivial because any arrangement will have at least two girls sitting next to each other. So we will change the question to at least two boys are next to each other. Let s count the opposite, i.e. no boys sitting next to each other. We will count the opposite. Fix a girl s sitting and arrange the other girls in the counterclockwise order. There are! 6 ways to arrange the girls. Between the girls there are 4 spots to insert the boys. There are 4 4 ways to insert the boys. Total number of ways to arrange them this way is 6 4 44. The total number of possible seating arrangement with a girl s seat fixed will be 6! 70 clockwise. 70-44=576. We need to divide this number by two because one seating has two different orientations (clockwise and 576 counter-clockwise). 88. Answer: 88 04,Huasong Yin