- #1
saadsarfraz
- 86
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Hi, how would you show that 4^(k)+4 * 9^(k) [tex]\equiv[/tex] 0 (mod 5)
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robert Ihnot said:Checking a few small values of k shows it is not always true.
d_leet said:I'm pretty sure that it is, in fact, always true.
NoMoreExams said:You know that
[tex] 4^k + 4 \cdot 9^k = 5k, k \in \mathbb{Z} [/tex]
wsalem said:Mentallic, you are on topic.
[tex]a \equiv b \mod m[/tex] is read "a is congruent to b modulo m". Mathematically, it means [tex]a - b = mk[/tex] (or a = mk +b) for a fixed [tex]m \in \mathbf{N}[/tex] and some [tex]k \in \mathbf{N}[/tex]. In other words, [tex]a-b[/tex] is divisible by [tex]m[/tex].
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