Advent calendar 2021
9 December
Eve writes down a sequence of consecutive positive integers (she writes more than one number). The sum of the numbers Eve has written down is 844.
Today's number is the smallest integer that Eve has written down.
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If Eve had written down an even number of numbers, then the mean of her numbers would be halfway between the two middle integers, so would be an integer and a half.
If Eve had written down two numbers, then the mean of those two numbers would be 422. This is not an integer and a half.
If Eve had written down three numbers, then the mean of those three numbers would be 281.3333. This is not an integer.
If Eve had written down four numbers, then the mean of those two numbers would be 211. This is not an integer and a half.
If Eve had written down five numbers, then the mean of those three numbers would be 168.8. This is not an integer.
If Eve had written down six numbers, then the mean of those two numbers would be 140.6666 This is an not integer and a half.
If Eve had written down seven numbers, then the mean of those two numbers would be 120.5714. This is an not integer.
If Eve had written down eight numbers, then the mean of those two numbers would be 105.5. This is an integer and a half, so this is possible.
The eight consecutive integers with a mean of 105.5 are 105 and 106, 104 and 107, 103 and 108, and 102 and 109.
To be certain that this is the only answer, we should rule out Eve having written more than eight numbers.
The sum of the numbers from 1 to 41 is 861. This is greater than 844, so Eve must have written fewer than 41 numbers.
If Eve had written down an odd number of numbers, then the mean of her numbers would be an integer. This means that the number of numbers would be a factor of the sum of all the numbers.
844's prime factorisation is 2×2×211, and so its only odd factor is 211. This is greater than 41, so Eve didn't write down this many numbers.
Eve therefore must have written down an even number of numbers.
If Eve wrote down an even number of numbers, then the mean of her numbers would be an integer plus a half. This implies that half the number of numbers would be a factor of the sum, and
the number of numbers cannot be a factor of the sum. This is only possible if there are 8 numbers.