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The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Answers (1)
Let 13a & 13b be the two numbers
13a*13b =2028
ab =12
(1,12) & (3,4) are the co prime no with product 12
Hence required numbers are (13*1, 13*12) & (13*3 , 13*4) So there are 2 such pairs.
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