From Newsgroup: rec.puzzles
On Thu, 30 Oct 2025 10:24:38 -0400, leflynn wrote:
A recent puzzle in Games Magazine asked readers to find two or more consecutive whole numbers that summed to a specified whole number. This brought the following problem to mind: Describe the set of whole numbers which do not admit such a solution.
L. Flynn
Thanks - an interesting question. I've never attempted a mathematical
proof before, so would welcome improvements and corrections.
SPOILER BELOW.
POILER BELOW.
OILER BELOW.
ILER BELOW.
LER BELOW.
ER BELOW.
R BELOW.
BELOW.
BELOW.
ELOW.
LOW.
OW.
W.
.
Problem: Describe the set of whole numbers which cannot be formed by
summing two or more consecutive whole numbers.
The sums of n consecutive numbers, starting at 1, 2, 3, ... is as shown
below.
Sum of 2 numbers: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ...
Sum of 3 numbers: 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, ...
Sum of 4 numbers: 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, ...
Sum of 5 numbers: 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, ...
Sum of 6 numbers: 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, ...
Sum of 7 numbers: 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, ...
Sum of 8 numbers: 36, 44, 52, 60, 68, 76, 84, 92, 100, 108, ...
Sum of 9 numbers: 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, ...
Sum of 10 numbers: 55, 65, 75, 85, 95, 105, 115, 125, 135, 145, ...
Of note, the first term in each row (3, 6, 10, 15 etc.) are the triangular numbers, and each sum in the row increases by the number of consecutive numbers under consideration.
Looking at the numbers included across all sequences, in ascending order,
we have 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17 ...
The numbers excluded are 2, 4, 8, 16 ...
So the theorem is: the set of numbers 2^p where p >= 0 cannot be formed by summing two or more consecutive whole numbers.
Proof: Given an arithmetic series of m consecutive integers starting at k
and with a common difference 1, we can express the series as follows:
k, k + 1, k + 2 ... k + m - 2, k + m - 1
We can calculate the sum (S) of the series:
S = k + (k + 1) + (k + 2) + ... (k + m - 2) + (k + m - 1)
Reversing the order of the terms gives:
S = (k + m - 1) + (k + m - 2) + ... (k + 2) + (k + 1) + k
Adding we get:
2S = m(2k + m - 1)
S = m(2k + m - 1)/2
We note that 2^p only has even factors, while the sum of consecutive
numbers (S) has the following factors:
m odd, factors: odd and even / 2.
m even, factors: even / 2 and odd.
In each case the number formed by summing two or more consecutive whole numbers has at least one odd factor. Therefore the set of numbers 2^p,
where p >= 1, cannot be formed by summing two or more consecutive whole numbers.
I still need to show that all numbers with an odd factor can be decomposed into the sum of consecutive numbers. I'll think about that.
--
David Entwistle
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