rx10 wrote:
This can be done with trying out options.
The LEAST value of \(c\) for which both \(a\) & \(b\) will be integers as given in the stem.
Answer C
Hello!
can you please demonstrate your work? how you "tried out the options"? If a, b, and c are all positive integers, then (by the integer multiplication closure property (google "proofwiki integer multiplication is closed")) a*b, a*c, and b*c should all be positive integers as well.
I'm thinking that c cannot be an integer if a & b are positive integers and a = 75/(10000*c) and b = 9/(100*c)
kumarneupane4344 wrote:
If \(a*c=0.0075\) and \(b*c=0.09\),where a,b and c are positive integers. What is the least possible value of c?
A. 100
B. 200
C. 400
D. 800
E. Can not be determined.
Kudos for the right solution.
If a and b are positive integers, and a = 3/(2^4 * 5^2 * c) and b = 3^2 / (2^2 * 5^2 * c), then the greatest possible value of c would be: c = (3 * 2^(-4) * 5^(-2)) = 3/400:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (3 * 2^(-4) * 5^(-2))) = 1
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (3 * 2^(-4) * 5^(-2))) = 3 * 2^2 = 12
The least possible value of c would be (k is any positive integer): c = (2^(-4) * 5^(-2) * k^(-1)) = 1/(400 * k):
-> let k = 1, c = 1/400:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (2^(-4) * 5^(-2) * 1^(-1))) = 3 * 1 = 3
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (2^(-4) * 5^(-2) * 1^(-1))) = 3^2 * 2^2 * 1 = 36
-> let k = 2, c = 1/800:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (2^(-4) * 5^(-2) * 2^(-1))) = 3 * 2 = 6
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (2^(-4) * 5^(-2) * 2^(-1))) = 3^2 * 2^2 * 2 = 72
-> let k = 3, c = 1/1200:
a = 3/(2^4 * 5^2 * c) = 3/(2^4 * 5^2 * (2^(-4) * 5^(-2) * 3^(-1))) = 3 * 3 = 9
b = 3^2 / (2^2 * 5^2 * c) = 3^2 / (2^2 * 5^2 * (2^(-4) * 5^(-2) * 3^(-1))) = 3^2 * 2^2 * 3 = 108
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so because k can be any positive integer, and as k increases from 1, c decreases from 1/(400*1), the least value of c that makes a & b positive integers and a = 75/(10000*c) and b = 9/(100*c) cannot be determined.