Two numbers, both greater than 29,have HCF 29 and LCM 4147. The sum of the numbers is:
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Solution:-
given by:- HCF 29 and LCM 4147
》Product of numbers = 29 × 4147.
》Let the numbers be 29a and 29b.
》Then, 29a × 29b = (24 ×4147)
》=> ab = 143.
》Now, co-primes with product 143 are (1, 143) and (11, 13).
》So,the numbers are (29 × 1, 29 × 143) and (29 × 11, 29 × 13).
》Since both numbers are greater than 29, the suitable pair is (29 × 11, 29 × 13)
》i.e., (319, 377).
》 Required sum = (319 + 377) = 696.
☆i hope its help☆
given by:- HCF 29 and LCM 4147
》Product of numbers = 29 × 4147.
》Let the numbers be 29a and 29b.
》Then, 29a × 29b = (24 ×4147)
》=> ab = 143.
》Now, co-primes with product 143 are (1, 143) and (11, 13).
》So,the numbers are (29 × 1, 29 × 143) and (29 × 11, 29 × 13).
》Since both numbers are greater than 29, the suitable pair is (29 × 11, 29 × 13)
》i.e., (319, 377).
》 Required sum = (319 + 377) = 696.
☆i hope its help☆
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hope you will understand that...thnx
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