The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A. 3 B. 4 C. 2 D. 1
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The answer is c)2br / Given:br / The product of two numbers is 2028.br / i.e. XY=2028br / The prime factorisation of 2028 are 2,2,3,13,13.br / The hcf of two number is 13.br / X=13(A) Y=13(B)br / From the prime factorisation ,the combinations formed are br / 1)X=13(1) Y=13(2*2*3)br / 2)X=13(2) Y=13(2*3)br / 3)X=13(2*2) Y=13(3)br / 1) AND 3) have the HCF 13.2)have HCF 26.br /strong Hence there are totally 2 pairs available.(13,2028),(52,39)/strongbr /
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