Manhattan Challenge Problem of the week! - 04/23/07

Which of the following sets includes ALL of the solutions of x that will satisfy the equation:

Answer: OA - C

OE - One way to solve equations with absolute values is to solve for x over a series of intervals. In each interval of x, the sign of the expressions within each pair of absolute value indicators does not change.

In the equation Ix-2I - Ix-3I = Ix-5I , there are 4 intervals of interest:

x less than 2 - In this interval, the value inside each of the three absolute value expressions is negative.

2 less than x less than 3 - In this interval, the value inside the first absolute value expression is positive, while the value inside the other two absolute value expressions is negative.

3 less than x less than 5 - In this interval, the value inside the first two absolute value expressions is positive, while the value inside the last absolute value expression is negative.

5 less than x - In this interval, the value inside each of the three absolute value expressions is positive.

Use each interval for x to rewrite the equation so that it can be evaluated without absolute value signs.

For the first interval, x less than 2, we can solve the equation by rewriting each of the expressions inside the absolute value signs as negative (and thereby remove the absolute value signs):

- x + 2 - ( -x + 3 ) = - x + 5

- x + 2 + x - 3 = - x + 5

x = 6

Notice that the solution x = 6 is NOT a valid solution since it lies outside the interval x less than 2. (Remember, we are solving the equation for x SUCH THAT x is within the interval of interest).

For the second interval 2 less than x less than 3, we can solve the equation by rewriting the expression inside the first absolute value sign as positive and by rewriting the expressions inside the other absolute values signs as negative:

x - 2 - ( -x + 3 ) = - x + 5

x - 2 + x - 3 = - x + 5

3x = 10

x = 10/3

Notice, again, that the solution x = 10/3 is NOT a valid solution since it lies outside the interval 2 less than x less than 3

For the third interval 3 less than x less than 5, we can solve the equation by rewriting the expressions inside the first two absolute value signs as positive and by rewriting the expression inside the last absolute value sign as negative:

x - 2 - ( x - 3 ) = - x + 5

x - 2 - x + 3 = - x + 5

x = 4

The solution x = 4 is a valid solution since it lies within the interval 3 less than x less than 5

Finally, for the fourth interval 5 less than x, we can solve the equation by rewriting each of the expressions inside the absolute value signs as positive:

x - 2 - ( x - 3 ) = x - 5

x - 2 - x + 3 = x - 5

x = 6

The solution x = 6 is a valid solution since it lies within the interval 5 less than x.

We conclude that the only two solutions of the original equation are x = 4 and x = 6. Only answer choice C contains all of the solutions, both 4 and 6, as part of its set.

Therefore, C is the correct answer.

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.

Problem Solving - 21

Mrs. Napueta has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Napueta been given if there are 120 ways that she could distribute the vouchers?

(A) 13

(B) 14

(C) 15

(D) 16

(E) more than 16


Answer: C

Formula - Number of ways of distributing n identical things among r persons when each person may get any number of things = C(n+r-1, r-1)

Hence

Let total number of vouchers = n

All 4 nephews get atleast 2 vouchers, thus remaining vouchers to be distributed among 4 nephews where each can get 0 or more vouchers = n-8

Using the formula above we get = C(n-8+4-1, 3) = (n-8+4-1)C3 = 120 = (n-5)(n-6)(n-7) = 120 * 3! = 720

Hence n = 15

DS Question - 26

Is 5^k less than 1,000?

(1) 5^(k-1) greater than 3,000

(2) 5^(k-1) = 5^k - 500


Answer: D

From statement (1) - 5^(k-1) is greater than 3000

=> 5^k/5 is greater than 3000
=> 5^k is greater than 15000

Hence sufficient

From statement (2) - 5^(k-1) = 5^k - 500

=> 5^(k-1) = 5^k - 500
=> 5^k - 5^(k-1) = 500
=> 5^k(1- 5^-1) = 500
=> 5^k(4/5) = 500
=> 5^k = 2500/4 = 625
Therefore 5^k is less than 1000

Hence sufficient