Докажите, что для всех целых значений n выражение n^3-31n делится на 6. Допустим, что t - это остаток от деления
Докажите, что для всех целых значений n выражение n^3-31n делится на 6. Допустим, что t - это остаток от деления n на 6, то есть n = 6k + t. Или n^3 - 31n = (6k + t)^3 - 31(6k + t) = ? Если t = 1, то t^3 - 31t = ? Если t = 2, то t^3 - 31t = ? Если t = 3, то t^3 - 31t = ? Если t = 4, то t^3 - 31t = ? Если t = 5, то t^3 - 31t = ?
Золотой_Король 22
= 5^3 - 31(6k + 5)= 125 - 186k - 155
= -155 - 186k
Now, let"s consider each possible value of t and find the value of \(t^3 - 31t\):
1) If t = 1:
\(t^3 - 31t = 1^3 - 31(1) = 1 - 31 = -30\)
As we can see, -30 is divisible by 6, since -30 ÷ 6 = -5.
2) If t = 2:
\(t^3 - 31t = 2^3 - 31(2) = 8 - 62 = -54\)
Again, we see that -54 is divisible by 6, as -54 ÷ 6 = -9.
3) If t = 3:
\(t^3 - 31t = 3^3 - 31(3) = 27 - 93 = -66\)
Similarly, -66 is divisible by 6, as -66 ÷ 6 = -11.
4) If t = 4:
\(t^3 - 31t = 4^3 - 31(4) = 64 - 124 = -60\)
Once again, -60 is divisible by 6, as -60 ÷ 6 = -10.
5) If t = 5:
\(t^3 - 31t = 5^3 - 31(5) = 125 - 155 = -30\)
As we can see, -30 is divisible by 6, as -30 ÷ 6 = -5.
In each case, we have shown that \(t^3 - 31t\) is divisible by 6. Since t represents the remainder when n is divided by 6, this means that for any integer value of n, \(n^3 - 31n\) is divisible by 6. Thus, the expression holds true for all integer values of n.