R&N is a new candy with magical (yet mathematical) properties. Not all candies are the same size. When removing the largest R&N from a new bag, the bag becomes 1/2 full; removing the second largest makes the bg 1/3 full; removing the third largest makes it 1/4 full, and so on. Dr. Lutinski opens a new bag on Monday and eats the largest candy. On Tuesday he eats the next few largest, leaving 5/14 fewer than on Monday. How many did Dr. Lutinski eat?
Ya got me
On Monday there was 1/2 bag left.
That means 7/14 are left. If he is to eat a few more, then that means there are only 7/14 - 5/14 = 2/14 left after Tuesday's chow-down.
But, the next largest candy leaves 1/3 of the bag full. 1/3 > 2/14, so eh?
is that 1/3 of the original amount, or 1/3 of the starting amount on Tuesday?
Does "5/14 fewer" mean 5/14 of the original amount, or 5/14 of the 1/2-bag at the dawn of Tuesday?
To solve this problem, we need to determine the number of candies eaten by Dr. Lutinski. Let's break down the problem step by step:
Step 1: Determine the number of candies in the bag on Monday.
Let's assume that there are N candies in the bag on Monday.
Step 2: Determine the number of candies in the bag on Tuesday.
According to the problem, when Dr. Lutinski eats the largest candy, the bag becomes 1/2 full. This means that after eating the largest candy, there are N/2 candies left in the bag.
Step 3: Calculate the number of candies left in the bag after Dr. Lutinski eats the next few largest.
If after eating the largest candy (N/2 candies), he eats the next few largest, leaving 5/14 fewer candies than on Monday, we can set up the following equation:
N/2 - (5/14) = N - x
Where x represents the number of candies eaten by Dr. Lutinski. Rearranging the equation:
x = N - (N/2 - 5/14)
x = N - N/2 + 5/14
x = (14N - 7N + 5)/14
x = (7N + 5)/14
Step 4: Find the value of N that satisfies the equation.
To find the value of N, we need to solve for a whole number that satisfies the equation (7N + 5)/14.
For this equation to have a whole number solution, the numerator (7N + 5) must be divisible by the denominator (14).
Since the numerator (7N + 5) is always odd (resulting in an odd number + 5), the numerator must be divisible by 7 to obtain a whole number solution.
Let's try different values of N until we find a solution that satisfies the equation.
N = 7 (Try a multiple of 7)
x = (7 * 7 + 5)/14 = 49/14 = 3.5 (Not a whole number)
N = 14 (Try a larger multiple of 7)
x = (7 * 14 + 5)/14 = 99/14 = 7.0714 (Not a whole number)
N = 21 (Try an even larger multiple of 7)
x = (7 * 21 + 5)/14 = 154/14 = 11 (Whole number!)
Therefore, Dr. Lutinski ate 11 candies from the bag.
Please note that this solution assumes that each candy size is unique and there are no ties in size between candies.