What is the maximum number of zeroes that can appear at the end of the integer X=1^x+2^x+3^x+4^x (where x>0)

Answer 1

Probably two. If x = 3, then you get 1 + 8 + 27 + 64 = 100.

Yes, it can't be more than two. This can be shown with some mod 8 arithmetic. If x > 3, then:

1^x = 1 (mod 8)
2^x = 0 (mod 8)
3^x = 3 (mod 8) or 1 (mod 8)
4^x = 0 (mod 8)

So, the sum of the four numbers is either 2 (mod 8) or 4 (mod 8), indicating it cannot be divisible by 8. And, if the number is not divisible by 8, it cannot be divisible by 1000 or any higher power of 10.

Interesting question...

Answer 2

I liked this question too. I gave you a stern

I will think of it before I read other answers, seems to be a good quiz

Ana

EDIT

1^x = 1 forall x, Im guessing here that x is a positive integer

Anyway, if x = 0, then 1^x + 2^x + 3^x + 4^x = 4, so, it wont have any zeros.

If x = -1, then 1 + 1/2 + 1/3 * 1/4 wont have any zeros, either. And 2^x + 3^x +4^x will decrease is x is more a more negative integer. It will tend to 0, but it will never get to be 0. So, 1^x + 2^x + 3^x + 4^x will never be 0 , either.

What if x is a rational number?

1 + sqrt 2 + sqrt 3 + sqrt 4 approx 1+ 1.4 + 1.7 + 2. It wont be zero, either. It wont be 0 if x is a fraction 1/n, where n is a natural number, n different from 0. If x = -1/n, we will have a sum of 1 + 4 numbers, none of which will be greater than 0, no solution, either.

What about a fraction m/n? This could eventually work. I will study this possibility later.

But now I want to work with x, when its a positive integer.

1 + 2^2 + 3^2 + 4^2 = 1 + 2^2 + 2^4 + 3^2

1 + 3^2 = 10

And 2^2 (1+2^2) = 2^2 (5) = 20, so the result is 30, it has one zero.

I will go on thinking about this question, but now I have to work

Ana

Answer 3

2x^3 / (-3x^2-4x-4) lim x tending to infiniti to resolve such concern constantly divide the numerator and denominator via optimal means of x interior the equation, here its x^3. dividing numerator and denominator via x^3 we get; -2 / ( (3/x) + (4/x^2) + (4/x^3) ) here as x has a tendency to infinity 3/x has a tendency to 0, 4/x^2 has a tendency to 0 , and four/x^3 has a tendency to 0. so denominater has a tendency to 0. as a effect answer is -2/0 this is minus infiniti

Answer 4

None.

The only way to get 0 by multiplying non-zero numbers is to multiply an even number by 5. You could have lots of even numbers, but there are no 5s.

Answer 5

If you put x infinite then U'll get infinite 0s.