x=(2^n + 3^n + 5^n + 7^n) / 11^n
n is a natural number........ Find all the possible integer value of x

Question

x=(2^n + 3^n + 5^n + 7^n) / 11^n
n is a natural number........ Find all the possible integer value of x

experimentx · Answer

http://www.wolframalpha.com/input/?i=%282^n+%2B+3^n+%2B+5^n+%2B+7^n%29+where+n%3D1%2C2%2C3%2C4%2C5%2C6 http://www.wolframalpha.com/input/?i=%2811^n%29+where+n%3D1%2C2%2C3%2C4%2C5%2C6 for n>1 the x<1

experimentx · Answer

n=0 and n->inf might be only case when x can have integer value.

anonymous · Answer

n cannot be zero as n is a natural number

anonymous · Answer

Actually there is a nice solution......

experimentx · Answer

could you check your question again? http://www.wolframalpha.com/input/?i=plot+%282%5En+%2B+3%5En+%2B+5%5En+%2B+7%5En%29+%2F+11%5En+from+n%3D1+to+6 the only possible value x can have is 1 and n is not a natural number at that point.

anonymous · Answer

Actually, thats the solution

anonymous · Answer

That website looks great..... I proved there is no solution(1 page proof)..... And it just using range

anonymous · Answer

Hmm. How about this.
In order for it to be an integer the numerator must be divisible by 11.
Units place of any multiple of 11 is always 1.
Units place of numerator can be.
|dw:1347535741594:dw|
Here I've written down all the possible unit places. Note that 1 is not one of them. Hence not divisible. Is this correct?