Question

Let (z) be a positive integer such that the last four digits of (z^31) are 0001. Show that the last four digits of (z) are 0001.

a) (z) is a prime number.
b) (z) is an even number.
c) (z) is a μltiple of 5.
d) (z) is a perfect square.

Answer 1

To show that the last four digits of z are 0001 when the last four digits of z^31 are 0001, we can consider the possible values of z. None of the given options are necessary to prove this.

Step-by-step explanation:

To show that the last four digits of z are 0001 when the last four digits of z^31 are 0001, we can consider the possible values of z. Since the last four digits of z^31 are 0001, z^31 ends in 1. This means that the last digit of z must be either 1 or 9. However, if z ends in 9, then z^31 will end in 9 as well, not 1. Therefore, the last digit of z must be 1.

None of the given options (a) (z is a prime number), (b) (z is an even number), (c) (z is a multiple of 5), or (d) (z is a perfect square) are necessary to prove that the last four digits of z are 0001.

Answer 2

To prove that (z) ends with 0001 given that (z^31) ends with 0001, we use modular arithmetic and properties of coprime numbers with respect to 10,000. Z must be odd and coprime with 10,000; hence, cannot be a multiple of 2 or 5. By contradiction, z must end with 0001, while being a prime or a perfect square is not necessarily implied.

Step-by-step explanation:

The question requires us to prove that if (z^31) ends with the digits 0001, then (z) must also end with 0001. Starting with the given that (z^31) has 0001 as the last four digits, let's consider the properties of numbers and their powers modulo 10,000 (since we are only interested in the last four digits).

For (z^31) to end with 0001, z must be odd because even numbers powered to any integer will result in a number that has 0 as the last digit. Furthermore, as z^31 ≡ 1 (mod 10,000), z must be coprime with 10,000 meaning it cannot have any prime factors in common with 10,000 (which are 2 and 5). Thus, z cannot be a multiple of 2 or a multiple of 5 (which rules out b and c).

Now, consider if (z) was not ending with 0001. Raising such a (z) to an odd power like 31 would yield a result that does not end with 0001. Hence, by contradiction, z must end with 0001. The property that (z) ends with 0001 does not inherently mean that (z) is a prime or a perfect square (which rules out a and d). With this, we conclusively demonstrate that for (z^31) to end with 0001, (z) must end with 0001.