(A) Three distinct positive integers belong to a
harmonic sequence. One of these integers equals the product of the other two, and their sum is 1983. Determine the three integers.
(B) Determine three distinct positive integers, belonging to a harmonic sequence, whose product is 3600.
|
|
Submitted by K Sengupta
|
|
Rating: 4.0000 (1 votes)
|
|
|
Solution:
|
(Hide)
|
PART (A):
Let the three positive integers be a,b and c with a< b< c
Then, b = 2ac/(a+c); and, c=ab(given); so that:
2*a^2*c/(a+c)= c; giving, c=a(2a-1); so that:
b = 2a-1
Accordingly, a+b+c = 1983 gives:
a^2 +a -992 = 0; giving a = 31(ignoring the negative root)
Consequently, b = 31*2 – 1 = 61 and c =a*b = 1891
Hence the required integers are 31, 61 and 1891.
PART(B)
Let the three positive integers be a,b and c with a< b< c
Then, b = 2ac/(a+c); and, abc=3600(given); so that:
a*c*(2ac/(a+c)) = 3600
or, ac = 30t, where t^2 = 2(a+c)
Now, t^2 = 2(a+c) gives:
t^2 = 2a + 60*t/a
or, 2*a^2 – a*t^2 +60t =0
or, a = (t^2 +/- sqrt(t^4 – 480t))/4
Since a is a positive integer, it follows that (t^4 – 480t) must be a perfect square.
By inspection, we observe that (t^4 – 480t) is a perfect square for t=8, since:
8^4 – 480*8 = 256 = 16^2
But t =8, gives a = (64 +/- 16)/4 = 20, 12, giving:
c = 30*t/a = 12, 20 and b = 2ac/(a+c) = 15
Hence the required integers are either 12, 15, 20 or; 20, 15, 12.
|
