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Question 145
The product of two numbers is 6760 and their HCF is 13. How many such pair of numbers can be formed?
Solution
Let the two numbers be 13x and 13y since HCF is divisible by two numbers.
13x $$\times$$ 13y = 6760
169xy = 6760
xy = 40
For HCF to be 13, there should be no common factors except 1.
Then, (x,y) can be (5,8) and (1,40).
Therefore, The numbers can be 13*5,13*8 = 65,104 and 13*1,13*40 = 13,520
Therefore, There can be 2 pairs of such numbers.
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