Question 1176058: Yann writes down the first n consecutive positive integers, 1, 2, 3, 4, . . . , n − 1, n. He removes four different integers p, q, r, s from the list. At least three of p, q, r, s are consecutive and 100 < p < q < r < s. The average of the integers remaining in the list is 89.5625. The number of possible values of s is
answer options: 30,32,28,24,26
Answer by greenestamps(13198)   (Show Source):
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The average of the remaining (n-4) integers is 89.5625 = 89 9/16.
That means n-4 is a multiple of 16.
The sum of the remaining (n-4) numbers is approximately 90(n-4), and four integers greater than 100 were removed from the list. So the average of the n integers is slightly more than 90. Since the list is the consecutive integers starting with 1, that means the number of numbers in the list, n, is about 180.
n-4 is a multiple of 16, so n is 4 more than a multiple of 16. The numbers close to 180 that are 4 more than a multiple of 16 are 164, 180, and 196. So it is almost certain that n is 180.
The sum of the first n=180 positive integers is (180)(181)/2 = 16290.
The sum of the n-4=176 integers remaining in the list after p, q, r, and s are removed is 176(89.5625) = 15763.
So the sum of p, q, r, and s is 16290-15763 = 527.
p, q, r, and s are all greater than 100; at least three of them are consecutive; and their sum is 527. The possibilities are....
101, 102, 103, 221
102, 103, 104, 218
103, 104, 105, 215
...
129, 130, 131, 137
130, 131, 132, 134
That is 30 different combinations, with 30 different values of s.
ANSWER: The number of possible values of s is 30.
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