There are fourteen positive integers less than 1000 out of which 2 have equal values, 3 others also have equal values, 4 others have equal values and the remaining 5 are distinct. If one of these integers is equal to the average of all these ten integers and is less than 50, then what can be the largest possible value of one of the ten integers?
Explanation:
Let the numbers be - a, a, b, b, b, c, c, c, c, d, e, f, g and h.
Since we need to maximise one of these numbers, we need to minimise others and maximise the average.
∴ Let us assume c = 1, b = 2, a = 3, d = 4, e = 5, f = 6, g = average of all the numbers and h is the highest number.
Since average is less than 50, highest value of g is 49. ∴ Sum of all 14 numbers = 14 × 49 = 686
⇒ 686 = 3 + 3 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 4 + 5 + 6 + 49 + h
⇒ h = 606
Hence, 606.
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