a deck of 150 cards has cards numbered 1,2,3,...,150. the deck is shuffled and dealt into 5 different piles. what is the number of different sums possible for cards in any one pile?
a) 2325
b) 2600
c) 3601
d) 4065
e) 4066
To find the number of different sums possible for cards in any one pile, we need to analyze the problem and come up with a strategy.
Given that there are 150 cards numbered from 1 to 150, and they are shuffled and dealt into 5 different piles, let's consider the possibilities for the minimum and maximum sums that can be obtained in a pile.
The minimum sum would occur if we had the first 30 cards (numbers 1 to 30) in one pile. The sum of this sequence can be calculated using the formula for the sum of consecutive numbers: \( \text{{sum}} = \frac{{n \times (n + 1)}}{2} \), where \( n \) is the number of terms. In this case, \( n = 30 \). Plugging these values into the formula, we get: \( \frac{{30 \times (30 + 1)}}{2} = 465 \) as the minimum sum.
The maximum sum would occur if we had the last 30 cards (numbers 121 to 150) in one pile. Using the same formula, we find that the maximum sum is \( \frac{{30 \times (120 + 151)}}{2} = 4335 \).
Now, we need to determine the number of different sums between the minimum and maximum sums. This can be done by finding the number of integers in the range from 465 to 4335, inclusive. The formula to calculate the number of integers in a range is \( \text{{count}} = \text{{end}} - \text{{start}} + 1 \), where \(\text{{end}}\) is the maximum value and \(\text{{start}}\) is the minimum value. Plugging in the values, we get \( \text{{count}} = 4335 - 465 + 1 = 3871 \).
However, we need to subtract the number of piles, which is 5, since we cannot have duplicate sums in different piles. This gives us \( \text{{count}} = 3871 - 5 = 3866 \).
Therefore, the correct answer is closest to 3866, which is option e) 4066.