Let the required numbers be x, 2x, 3x. Then, their H.C.F =x. so, x= 12
Therefore, The numbers are 12, 24, 36
Product of numbers = 11 x 385 = 4235
Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35
Now, co-primes with product 35 are (1,35) and (5,7)
So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)
Since one number lies 75 and 125, the suitable pair is (55,77)
Hence , required number = 77
Let the numbers be 3x and 4x . Then their H.C.F = x. So, x=4
Therefore, The numbers are 12 and 16
L.C.M of 12 and 16 = 48
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = = = =
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
=>Required number = (90 x 4) + 4 = 364.
H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161.
Let the numbers be a and b . Then , ab= 161.
Now, co-primes with product 161 are (1, 161) and (7, 23).
Since x and y are prime numbers and x >y , we have x=23 and y=7.
Therefore, 3y-x = (3 x 7)-23 = -2
Let the numbers be 37a and 37b. Then , 37a x 37b =4107 => ab = 3.
Now co-primes with product 3 are (1,3)
So, the required numbers are (37 x 1, 37 x 3) i.e, (37,111).
Therefore, Greater number = 111.
Other number = = 308
60 is not a multiple of 8
L.C.M of 5, 6, 7, 8 = 840
Therefore, Required Number is of the form 840k+3.
Least value of k for which (840k+3) is divisible by 9 is k = 2
Therefore, Required Number = (840 x 2+3)=1683
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