David W. answered 11/13/16
Tutor
4.7
(90)
Experienced Prof
First, let’s find the LCM (least common multiple) of the values [ 3, 4, 5, 6, 8, 10 ]. We do that by listing the prime factors of each number:
3 : 3
4 : 2*2
5 : 5
6 : 2*3
8 : 2*2*2
10 : 2*5
Then, we select the minimum group of factors that contains all of those numbers:
2*2*2*3*5 = 120
That means that the pattern repeats each 120 numbers. Suddenly, it makes sense why there were 360 turkeys on the farm.
These turkey numbers do not have any of the listed problems:
1 2 7 11 13 14 17 19 22 23 26 29
31 34 37 38 41 43 46 47 49 53 58 59
61 62 67 71 73 74 77 79 82 83 86 89
91 94 97 98 101 103 106 107 109 113 118 119
Repeats for 120+n and for 240+n.
There are 48*3=144 good turkeys.
Now, to find the numbers between 1 and 120, create a kind of “Sieve of Eratosthenes,” but eliminate multiples of only the listed factors. Each row in your table could have 2*3*5=30 numbers:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 . . . 30
31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 . . . 60
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 . . . 90
91 92 93 94 95 96 97 97 99 100 101 102 103 104 105 106 107 . . . 120
X-out –
-- all the numbers in the column below 3 (they are all multiples of 3), the entire columns with 6 and with 9 and with 12 and with 15 and with 18 and with 21 and with 24 and with 27 and with 30 at the top (they are all multiples of 3; note: some of these turkeys have multiple problems because they are also a multiple of 6);
-- the numbers below 5 and the entire columns beginning with 10, 15, 20, 25 and 30 (oh, some of these also have multiple problems because they are also a multiple of 10);
-- that leaves every fourth to eliminate (note: 8 is a multiple of 4). Note that 4 is 2*2, so numbers in alternate rows below 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28 and 30 (now, some turkeys have even more problems) may be carefully x’ed-out.
3 : 3
4 : 2*2
5 : 5
6 : 2*3
8 : 2*2*2
10 : 2*5
Then, we select the minimum group of factors that contains all of those numbers:
2*2*2*3*5 = 120
That means that the pattern repeats each 120 numbers. Suddenly, it makes sense why there were 360 turkeys on the farm.
These turkey numbers do not have any of the listed problems:
1 2 7 11 13 14 17 19 22 23 26 29
31 34 37 38 41 43 46 47 49 53 58 59
61 62 67 71 73 74 77 79 82 83 86 89
91 94 97 98 101 103 106 107 109 113 118 119
Repeats for 120+n and for 240+n.
There are 48*3=144 good turkeys.
Now, to find the numbers between 1 and 120, create a kind of “Sieve of Eratosthenes,” but eliminate multiples of only the listed factors. Each row in your table could have 2*3*5=30 numbers:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 . . . 30
31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 . . . 60
61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 . . . 90
91 92 93 94 95 96 97 97 99 100 101 102 103 104 105 106 107 . . . 120
X-out –
-- all the numbers in the column below 3 (they are all multiples of 3), the entire columns with 6 and with 9 and with 12 and with 15 and with 18 and with 21 and with 24 and with 27 and with 30 at the top (they are all multiples of 3; note: some of these turkeys have multiple problems because they are also a multiple of 6);
-- the numbers below 5 and the entire columns beginning with 10, 15, 20, 25 and 30 (oh, some of these also have multiple problems because they are also a multiple of 10);
-- that leaves every fourth to eliminate (note: 8 is a multiple of 4). Note that 4 is 2*2, so numbers in alternate rows below 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28 and 30 (now, some turkeys have even more problems) may be carefully x’ed-out.
This leaves the numbers listed.