Show that `7^(2n-1) - 1` is divisible by 48

Asked on by mirex

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

It's not true for all `n,` or for any `n` for that matter. Try a few examples. For `n=1,` `7^(2n-1)-1=6,` which isn't divisible by `48.`

For `n=2,` `7^(2n-1)-1=7^3-1=342,` which again isn't divisible by `48.`

The real relationship is that `(7^(2n-1)-1)/48` always has fractional part `1/8.` This can be proved by induction, but first you should double check that you didn't misread or miscopy the problem.

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

My guess is that the problem was to prove that `7^(2n)-1` is divisible by `48.` In that case, the previous induction proof can easily be modified to prove this.

Another way would be to write

`7^(2n)-1=49^n-1=(48+1)^n-1` and then use the binomial theorem.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It has to be determined if `7^(2n - 1) - 1` is divisible by 48.

For n = 1, `7^(2n - 1) - 1 = 48`

Let `7^(2n - 1) - 1` be divisible for any integer n,

`7^(2n - 1) - 1 = 48*k` , where k is an integer

The value of the expression for n + 1 is:

`7^(2(n+1) - 1) - 1 `

= `7^((2n + 2) - 1) - 1`

= `7^(2n + 1) - 1`

= `7^(2n - 1)*7^2 - 1`

Substitute the expression derived for n

= `(48k +1)*49 - 1`

= `48*49*k + 49 - 1`

= `48*49*k + 48`

This proves that `7^(2n - 1) - 1` is divisible by 48 for all `n >= 1`

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