Question
Three pairwise co-prime natural
numbers have the following property: The product of the first and second numbers is 210, and the product of the second and third numbers is 154. If the sum of the smallest and the largest of these three numbers is 'p', find the number of factors of 'p'.Solution
ATQ, Let the numbers be 'a', 'b' and 'c' According to the question, ab = 210 bc = 154 Now, b = HCF of (210 and 154) 210 = 2 × 3 × 5 × 7 154 = 2 × 7 × 11 Common part = 2 × 7 = 14 Therefore, b = 14 Now, a = 210/14 = 15 c = 154/14 = 11 So the three numbers are: 15, 14, 11 Least number = 11 Greatest number = 15 Therefore, p = 11 + 15 = 26 Now, factors of 26 are: 1, 2, 13, 26 ∴ total number of factors of 26 = 4
elude
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