Question 1138467: Determine the number of ordered pairs of positive integers (a,b) such that a and b are both divisors of 1260 and a/b
Answer by greenestamps(13200)   (Show Source):
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The prime factorization of 1260 is 1260 = 2*2*3*3*5*7
One of the factors of 1260 is 20 = 2*2*5; 1260/20 = 63.
If a=20, then b can be any number of the form 20n, where n is any factor of 63.
63 = 3*3*7; the number of factors of 63 is (2+1)(1+1) = 3*2 = 6.
So there are 6 ordered pairs of positive integers (20,b) for which 20 and b are both divisors of 1260 and a divides b: (20,20), (20,60), (20,140), (20,180), (20,420), and (20,1260).
Perform a similar analysis for each value of a:"
+-------------------------------------------+
| | | | number of |
| | | | possible |
| | | factors | values |
| a | n=1260/a | of n | for b |
+-------------------------------------------+
1 1260 2*2*3*3*5*7 3*3*2*2 = 36
2 630 2*3*3*5*7 2*3*2*2 = 24
3 420 2*2*3*5*7 3*2*2*2 = 24
4 315 3*3*5*7 3*2*2 = 12
5 252 2*2*3*3*7 3*3*2 = 18
6 210 2*3*5*7 2*2*2*2 = 16
7 180 2*2*3*3*5 3*3*2 = 18
9 140 2*2*5*7 3*2*2 = 18
10 126 2*3*3*7 2*3*2 = 12
12 105 3*5*7 2*2*2 = 8
14 90 2*3*3*5 2*3*2 = 12
15 84 2*2*3*7 3*2*2 = 12
18 70 2*5*7 2*2*2 = 8
20 63 3*3*7 3*2 = 6
21 60 2*2*3*5 3*2*2 = 12
28 45 3*3*5 3*2 = 6
30 42 2*3*7 2*2*2 = 8
35 36 2*2*3*3 3*3 = 9
36 35 5*7 2*2 = 4
42 30 2*3*5 2*2*2 = 8
45 28 2*2*7 3*2 = 6
60 21 3*7 2*2 = 4
63 20 2*2*5 3*2 = 6
70 18 2*3*3 2*3 = 6
84 15 3*5 2*2 = 4
90 14 2*7 2*2 = 4
105 12 2*2*3 3*2 = 6
126 10 2*5 2*2 = 4
140 9 3*3 3 = 3
180 7 7 2 = 2
210 6 2*3 2*2 = 4
252 5 5 2 = 2
315 4 2*2 3 = 3
420 3 3 2 = 2
630 2 2 2 = 2
1260 1 = 1
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total: 330
ANSWER: There are 330 ordered pairs (a,b) in which a and b are both divisors of 1260 and a divides b.
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