Exponent

[yellowho]

(1+5^1000)/2^(X) is not integer. What is X?

Since 5 to anything has a unit digit of 5. Adding one will make it even so there's at least 1 2 in there. How do you determine the rest? All I know now is X is not 1.

[Rahul@gurome]

(1+5^1000)/2^(X) is not integer. What is X?.

There may be more than 1 value of x.
A proper question should ask the minimum integral value of x.

Now note that, 5 = (4 + 1) = (2² + 1)
Thus any power of 5 will be of the of the form (4n + 1), where n is any non-negative integer.

Thus, (any power of 5 + 1) will be of the form (4n + 2) = 2(2n + 1)

Hence, (1 + 5^1000) is = 2*(Some ODD integer)

Now, if (1 + 5^1000)/(2^x) is not an integer, 2^x must be greater than 2.
Hence, minimum possible value of x is 2.

[yellowho]

Jut to clarify these all works as X:

2,6,10, 14, 18 etc.........



[quote="Rahul@gurome"][quote="yellowho"](1+5^1000)/2^(X) is not integer. What is X?.[/qJust uote]

There may be more than 1 value of x.
A proper question should ask the minimum integral value of x.

Now note that, 5 = (4 + 1) = (2² + 1)
Thus any power of 5 will be of the of the form (4n + 1), where n is any non-negative integer.

Thus, (any power of 5 + 1) will be of the form (4n + 2) = 2(2n + 1)

Hence, (1 + 5^1000) is = 2*(Some ODD integer)

Now, if (1 + 5^1000)/(2^x) is not an integer, 2^x must be greater than 2.
Hence, minimum possible value of x is 2.[/quote]

[Rahul@gurome]

Jut to clarify these all works as X:

2,6,10, 14, 18 etc.........

More precisely, x can be any integer greater than 1.

[yellowho]

Oops. I meant to ask If that expression 2^x IS DIVISIBLE. Then its just 2 * a bunch of odd numbers? How do you know when it ends?

[quote="Rahul@gurome"][quote="yellowho"]Jut to clarify these all works as X:

2,6,10, 14, 18 etc.........[/quote]

More precisely, x can be any integer greater than 1.[/quote]