Question

将63分成若干个连续自然数的和,一共有多少种不同的方法?

Answer 1

设从X的N个连续自然数和=63
则有
(X + X+N-1)*N / 2 = 63
即
(N + 2X-1)*N = 126
因N + 2X-1、N必不同奇偶,且N+2X-1必大于N【指当X≥1时,当X=0时显然无解】
则将126拆成两个大于1的数的积
2 3 6 7 9
①126 = 63*2
N = 2
N + 2X-1 = 63 解得X = 31
有63 = 31+32
②126 = 42*3
N = 3
N + 2X-1 = 42 解得X = 20
有63 = 20+21+22
③126 = 21*6
N = 6
N + 2X-1 = 21 解得X = 8
有63 = 8+9+10+11+12+13
④126 = 18*7
N = 7
N + 2X-1 = 18 解得X = 6
有63 = 6+7+8+9+10+11+12
⑤126 = 14*9
N = 9
N + 2X-1 = 14 解得X = 3
有63 = 3+4+5+6+7+8+9+10+11
综上,共此5种