# 3n+1_conjecture 7, There is 88 that is the sum of consecutive natural numbers. 連続する自然数の和が88になる、その連続した自然数を求めてみましょう。

Before 3n + 1 conjecture,
let’s get used to handling integers and do a preparatory exercise.

Q: There is 88 that is the sum of consecutive natural numbers.
then, can you find each natural number?

3n + 1の前に、整数と有理数の性質を辿りましょう。
Q: 連続する自然数の和が88になる、その連続した自然数を求めてみましょう。

It defines k and nk as natural number.
Then, denote as expression below.
88 = n1 + ,,, + nk (n1 ,,, nk are consecutive natural numbers)

Transforming from the sum formula of the difference sequence,
it becomes as follows.
88 = n1 + (n1 + 1) + (n1 + 2) + ,,, + (n1 + k – 1)
88 = kn1 + Σi=1k-1 i
88 = k(n1 + (k/2) – (1/2)) — (1)

In order (1) in left side to be an integer,
it is a necessary condition that k is an odd number.
Therefore, it becomes as follows about k.
88 ≡ 0 mod k (k is an odd number)
therefore, k = 11.

Substitute this k = 11 into (1).
88 = 11(n1 + (11/2) – (1/2))
88 = 11(n1 + 5)
therefore, n1 = 3.

Therefore, the solution to be obtained is “3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13”.