Thursday, May 31, 2007

Manhattan Challenge Problem of the week! - 16/04/07

x and y are positive integers. , what is the value of xy?

(A) 48
(B) 36
(C) 24
(D) 18
(E) 12

Answer: OA - E

5^x - 5^y = 2^(y-1) * 5^(x-1)
Divide both sides by 5^(x-1)....we get
[5^x/ 5^(x-1)] - [5^y/ 5^(x-1)] = 2^(y-1)

Simplifying the first tem we get

5 - [5^x/ 5^(x-1)] = 2^(y-1)

Add [5^y/ 5^(x-1)] to both sides...we get

5 = 2^(y-1) + [5^y/ 5^(x-1)]

We know that 2^(y-1) and [5^y/ 5^(x-1)] are positive and each of them is less than 5

We can now list all the possibilities, by testing small integer values for y. We only have to test a few values because we know that 2^(y-1) must be less than 5, which means that y can only be 1, 2 or 3. (If y is 4, then 2^(y-1) would be 8, which is greater than 5.)

If y = 1, then 2^(y-1) = 2^(1-1) = 2^0 , which means that [5^y/ 5^(x-1)] must equal 4 (remember, the sum of the two terms must be equal to 5). However, since x and y are positive integers, there is no way to make [5^y/ 5^(x-1)] equal to 4.

If y = 2, then 2^(y-1) = 2^(2-1) = 2^1 = 2, which means that [5^y/ 5^(x-1)] must equal 3. However, since x and y are positive integers, there is no way to make [5^y/ 5^(x-1)] equal to 3.

If y = 3, then 2^(y-1) = 2^(3-1) = 2^2 = 4 , which means that cmust equal 1. This is possible. Using y = 3, we can solve for x as follows:

[5^y/ 5^(x-1)] = 1
=> [5^3/ 5^(x-1)] = 1
=> 5^3 = 5^(x-1)
=> 3 = x-1
=> x = 4

The only possible solution is y = 3 and x = 4. Therefore, xy = (4)(3) = 12. The correct answer is E.