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Unformatted text preview: Solutions to Homework 5, Math 55 Section 4.1 6. By the product rule, there are 4 · 6 = 24 such routes. 18. (a) The positive integers less than 1000 which are divisible by 7 are 7 · 1 , 7 · 2 , . . . , 7 · b 999 7 c . Therefore, b 999 7 c = 142 numbers less than 1000 are divisible by 7. (b) The numbers less than 1000 which are divisible by both 7 and 11 are the ones divisible by 77; therefore, there are b 999 77 c = 12 of these. Subtracting, 142 12 = 130 of the numbers divisible by 7 are not divisible by 11. (c) By the previous part, 12 are. (d) There are b 999 11 c = 90 multiples of 11 less than 1000. Therefore, by inclusionexclusion, 142 + 90 12 = 220 are divisible by either 7 or 11. (e) By part (b), there are 130 multiples of 7 not divisible by 11. Similarly, there are 90 12 = 78 multiples of 11 not divisible by 7. Therefore, 130 + 78 = 208 are divisible by exactly one of 7 and 11. (f) Since 220 are divisible by either 7 or 11 by part (d), 999 220 = 779 are divisible by neither 7 nor 11. (g) There are 9 such 1digit numbers. For a 2digit number, we can choose the first digit to be 1 through 9, then we have 9 choices for the second digit; thus, there are 9 · 9 = 81 such 2digit numbers. Similarly, there are 9 · 9 · 8 = 648 such 3digit numbers. In total, there are 9 + 81 + 648 = 738 such numbers less than 1000. (h) There are 4 such 1digit numbers. For a 2digit number, if we choose one of the 4 possible even digits for the first digit, there are 4 possibilities for the second digit; if we choose one of the 5 possible odd digits for the first digit, there are 5 possibilities for the second digit. Therefore, there are 4 · 4+5 · 5 = 41 such 2digit numbers. Similarly, for a threedigit number, if the first digit is even (4 choices), then there are 4 choices for the third digit, then 8 choices for the second digit. If the first digit is odd (5 choices), then there are 5 choices for the third digit, then 8 choices for the second digit. Therefore, there are 4 · 4 · 8 + 5 · 5 · 8 = 328 such 3digit numbers. In total, there are 4 + 41 + 328 = 373 such numbers less than 1000. 22. (a) Since there are 10 ways to choose the first digit, then 9 ways to choose the second, and so on, there are 10 · 9 · 8 · 7 = 5040 strings of four decimal digits which do not contain the same digit twice. (b) There are 10 ways to choose each of the first three digits, and 5 ways to choose the last digit as even. Therefore, there are 10 3 · 5 = 5000 such strings. (c) There are 4 ways to choose which digit is not a 9, then 9 ways to choose that digit. Therefore, there are 4 · 9 = 36 such strings. 40. Out of these bit strings, 2 5 = 32 start with two 0’s; 2 4 = 16 end with three 1’s; and 2 2 = 4 start with two 0’s and end with three 1’s. Therefore, by inclusionexclusion, 2 5 + 2 4 2 2 = 44 start with two 0’s or end with three 1’s....
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This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.
 Spring '08
 STRAIN
 Math

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