we know 2^3-1 = 7
so 2^(3n) - 1 is divisible by 2^3-1 or 7
now we need to check for numbers of the form 2^(3n+1) - 1 and 2^(3m+ 2) - 1
now 2^(3n) mod 7 = 1
so 2^(3n+1) mod 7 = 2
so 2^(3n+1) - 1 mod 7 = 2 - 1 = 1 not divisible by 7
2^(3n+2) - 1 mod 7 = 4 - 1 = 3 not divisible by 7
hence n is of the form 3k
The first is n=3 ... 2^3 -1 = 8-1 = 7 = 7*1
The second is n = 6 ... 2^6 - 1 = 63 = 7*9
The third is n = 9 ... 2^9 -1 = 512 = 7*73
When n is a multiple of 3 the number 2^n - 1 is divisible by 7.
Then the other values of n are: n=12, n=15, n=18, n=21, n=24, n= 27, n=30, n= 33 ... and so on. OK!
Note first that n = 3 satisfies the condition, since 2^3 - 1 = 7.
Now look at n = 3k for integers k >= 1. Then 2^(3k) - 1 is divisible by 7 for k = 1.
Now suppose that the statement is true for some k > 1, i.e., 2^(3k) - 1 = 7p for
some integer p. Then 2^(3k) = 7p + 1, and
2^(3(k+1)) - 1 = 2^(3k + 3) - 1 = 2^3 * 2^(3k) - 1 = 8*(7p + 1) - 1 = 8*7p + 7 =
7*(8p + 1), which is divisible by 7. So the statement is true for k+1, and thus
by induction we have that 2^n - 1 is divisible by 7 for all n = 3k for k >= 1,
i.e., for all positive multiples of 3. I still need to look at other values of n.
Edit: O.k., now look at integers n = 3k + 1 for k >= 1. Then
2^n - 1 = 2^(3k + 1) - 1 = 2*2^(3k) - 1 = 2*(2^(3k) - 1) + 1 = 2*7p + 1,
since from before we know that 2^(3k) - 1 = 7p for some integer p.
But 2*7p + 1 = 7*(2p) + 1 cannot be divisible by 7 since 7*(2p) is divisible by 7,
and we can't have two consecutive integers both be divisible by 7.
Now look at integers n = 3k + 2 for k >= 1. Then
2^n - 1 = 2^(3k + 2) - 1 = 2^2 * 2^(3k) - 1 = 4*(2^(3k) - 1) + 3 = 4*7p + 3
since from before 2^(3k) - 1 = 7p for some integer p.
But 4*7p + 3 = 7*(4p) + 3 cannot be divisible by 7 since 7*(4p) is divisible by 7,
and we cannot have two positive integers that differ by 3 both be divisible by 7.
In this way, by looking at the cases n = 3k, n = 3k + 1 and n = 3k + 2 for all
integers k >= 1 we have covered all positive integers >= 3. Now for n = 1
and n = 2 it is clear that 2^n - 1 is not divisible by 7, and thus we have now
covered all positive integers. So, finally, we can conclude that the set of all
positive integers for which 2^n - 1 is divisible by 7 is the set of all positive
multiples of 3, i.e., {n: n = 3k, k an integer, k >= 1}.
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we know 2^3-1 = 7
so 2^(3n) - 1 is divisible by 2^3-1 or 7
now we need to check for numbers of the form 2^(3n+1) - 1 and 2^(3m+ 2) - 1
now 2^(3n) mod 7 = 1
so 2^(3n+1) mod 7 = 2
so 2^(3n+1) - 1 mod 7 = 2 - 1 = 1 not divisible by 7
2^(3n+2) - 1 mod 7 = 4 - 1 = 3 not divisible by 7
hence n is of the form 3k
The first is n=3 ... 2^3 -1 = 8-1 = 7 = 7*1
The second is n = 6 ... 2^6 - 1 = 63 = 7*9
The third is n = 9 ... 2^9 -1 = 512 = 7*73
When n is a multiple of 3 the number 2^n - 1 is divisible by 7.
Then the other values of n are: n=12, n=15, n=18, n=21, n=24, n= 27, n=30, n= 33 ... and so on. OK!
Note first that n = 3 satisfies the condition, since 2^3 - 1 = 7.
Now look at n = 3k for integers k >= 1. Then 2^(3k) - 1 is divisible by 7 for k = 1.
Now suppose that the statement is true for some k > 1, i.e., 2^(3k) - 1 = 7p for
some integer p. Then 2^(3k) = 7p + 1, and
2^(3(k+1)) - 1 = 2^(3k + 3) - 1 = 2^3 * 2^(3k) - 1 = 8*(7p + 1) - 1 = 8*7p + 7 =
7*(8p + 1), which is divisible by 7. So the statement is true for k+1, and thus
by induction we have that 2^n - 1 is divisible by 7 for all n = 3k for k >= 1,
i.e., for all positive multiples of 3. I still need to look at other values of n.
Edit: O.k., now look at integers n = 3k + 1 for k >= 1. Then
2^n - 1 = 2^(3k + 1) - 1 = 2*2^(3k) - 1 = 2*(2^(3k) - 1) + 1 = 2*7p + 1,
since from before we know that 2^(3k) - 1 = 7p for some integer p.
But 2*7p + 1 = 7*(2p) + 1 cannot be divisible by 7 since 7*(2p) is divisible by 7,
and we can't have two consecutive integers both be divisible by 7.
Now look at integers n = 3k + 2 for k >= 1. Then
2^n - 1 = 2^(3k + 2) - 1 = 2^2 * 2^(3k) - 1 = 4*(2^(3k) - 1) + 3 = 4*7p + 3
since from before 2^(3k) - 1 = 7p for some integer p.
But 4*7p + 3 = 7*(4p) + 3 cannot be divisible by 7 since 7*(4p) is divisible by 7,
and we cannot have two positive integers that differ by 3 both be divisible by 7.
In this way, by looking at the cases n = 3k, n = 3k + 1 and n = 3k + 2 for all
integers k >= 1 we have covered all positive integers >= 3. Now for n = 1
and n = 2 it is clear that 2^n - 1 is not divisible by 7, and thus we have now
covered all positive integers. So, finally, we can conclude that the set of all
positive integers for which 2^n - 1 is divisible by 7 is the set of all positive
multiples of 3, i.e., {n: n = 3k, k an integer, k >= 1}.