Question

Доказать, что 18^4 + 52^3 + 86^4 + 14 делится на 17

Answer 1

18⁴ + 52³ + 86⁴ + 14 ≡ 1⁴ + 1³ + 1⁴ + 14 ≡ 17 ≡ 0 (mod 17)

запись a ≡ b (mod c) - число а дает такой же остаток как и b при делении на с

Answer 2

To prove that the expression $18^4 + 52^3 + 86^4 + 14$ is divisible by 17, we need to show that the expression leaves a remainder of 0 when divided by 17.

First, let's compute the individual terms of the expression:

$18^4 = 18 \times 18 \times 18 \times 18 = 104976$ $52^3 = 52 \times 52 \times 52 = 140608$ $86^4 = 86 \times 86 \times 86 \times 86 = 592974736$

Now, let's add up all the terms:

$18^4 + 52^3 + 86^4 + 14 = 104976 + 140608 + 592974736 + 14$

To make the calculation easier, we can simplify the expression modulo 17:

$18^4 \equiv 1 \pmod{17}$ (Fermat's Little Theorem) $52^3 \equiv 1 \pmod{17}$ (Fermat's Little Theorem) $86^4 \equiv 1 \pmod{17}$ (Fermat's Little Theorem)

Now, let's rewrite the expression modulo 17:

$18^4 + 52^3 + 86^4 + 14 \equiv 1 + 1 + 1 + 14 \pmod{17} = 17 \pmod{17} = 0$

Since the simplified expression leaves a remainder of 0 when divided by 17, we can conclude that $18^4 + 52^3 + 86^4 + 14$ is divisible by 17.