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5 gru 2020 · Mathematic induction is a tremendously useful proof technique and today we use it to prove that 7^n - 1 is divisible by 6. This is an exercise from Introductory Discrete Mathematics by V.K....
A number $\overline{a_1a_2\ldots a_n}$ is divisible by 7 if $\overline{a_1a_2\ldots a_{n-1}} - 2\times a_n$ is divisible by 7 too. The first two statements are very well known and quite easy to prove.
Divisibility by 6: The number should be divisible by both \(2\) and \(3\). Divisibility by 7: The absolute difference between twice the units digit and the number formed by the rest of the digits must be divisible by \(7\) (this process can be repeated for many times until we arrive at a sufficiently small number).
A number is divisible by if and only if the last digits of the number are divisible by . Thus, in particular, a number is divisible by 2 if and only if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8.
There are a couple of standard approaches. One is to use Fermat's little theorem, which says that if p is a prime number, then np − n is divisible by p for all n. Since 42 = 2 × 3 × 7, what we need to do is to check that 2, 3, and 7 divide n7 − n, no matter what n is. That 7 does is direct from Fermat's little theorem.
29 lut 2012 · Simple steps are needed to check if a number is divisible by 7. First, multiply the rightmost (unit) digit by 2, and then subtract the product from the remaining digits. If the difference is divisible by 7, then the number is divisible by 7. Example 1: Is 623 divisible by 7? 3 x 2 = 6. 62 – 6 = 56. 56 is divisible by 7, so 623 is divisible by 7.
Example 1: Use mathematical induction to prove that [latex]\large{n^2} + n[/latex] is divisible by [latex]\large{2}[/latex] for all positive integers [latex]\large{n}[/latex]. a) Basis step: show true for [latex]n=1[/latex].
