# HCF OF THREE NUMBERS IS 8,SUM OF THE NUMBERS IS 80, LST THE POSSIBLE SETS OF SUCH THREE NUMBERS. EXPLAIN BRIEFLY

We have HCF of three numbers =   8  , So these three numbers are  :

8x  , 8y  and 8 z   , Here  x  ,  y   and z are co - prime numbers

Also given Sum of these numbers is 80  , So

8x  + 8 y  + 8z =  80

8 (  x  + y  + z  ) =  80

x  + y  + z   = 10

Then  , The pair of numbers that are co-prime to each other and sum up to 10 are  (  1  ,  2  ,  7 )  , (  1 , 3 , 6 ) and  (  1  , 4  ,  5 )

Hence, only three pairs of such numbers are possible. The numbers are (   8 , 16 , 56 ) , ( 8 , 24 , 48 ) and ( 8 , 32 , 40 )                    ( Ans )

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