What is the number in the unit’s place of 999^22292582Options :9731

The number in the unit's place of 999^22292582 is 1.

Here's the step-by-step explanation:

1. First, we need to understand the pattern of the last digit of 999 raised to different powers.

2. If we look at the powers of 9, we see a pattern:

   • 9^1 ends in 9
   • 9^2 ends in 1
   • 9^3 ends in 9
   • 9^4 ends in 1
   • and so on...

3. So, the pattern repeats every two powers.

4. Now, we need to find the remainder when 22292582 is divided by 2 (since the pattern repeats every two powers).

5. The remainder is 0, which means 22292582 is an even number.

6. Therefore, 999^22292582 will end in the same digit as 9^2, which is 1.

So, the number in the unit's place of 999^22292582 is 1.

The number in the unit's place of 999^22292582 is 1.

Here's the step-by-step explanation:

1. First, we need to understand the cyclicity of the last digit. The number 9 has a cyclicity of 2, which means the last digit repeats every 2 powers. The cycle is {9, 1}.

2. To find out which number in the cycle is the last digit of 999^22292582, we need to find the remainder of the exponent (22292582) when divided by the cyclicity (2).

3. 22292582 divided by 2 gives a remainder of 0.

4. When the remainder is 0, we take the last number in the cycle, which is 1.

So, the number in the unit's place of 999^22292582 is 1.

The number in the unit's place of 999^22292582 is 1.

Here's the step-by-step explanation:

1. First, we need to understand the cyclicity of 9. The cyclicity of 9 is 2, which means the last digit repeats every 2 powers. The cycle of 9 is (9,1).

2. Then, we need to find the remainder when the power (22292582) is divided by the cyclicity (2). The remainder is 0.

3. Since the remainder is 0, we take the last number in the cycle of 9, which is 1.

So, the number in the unit’s place of 999^22292582 is 1.

The number in the unit's place of 999^22292582 is 1.

Here's the step-by-step explanation:

1. First, we need to understand the cyclicity of 9. The cyclicity of 9 is 2, which means the last digit repeats every 2 powers. The cycle of 9 is (9,1).

2. Then, we need to find the remainder when the power (22292582) is divided by the cyclicity (2). The remainder is 0.

3. Since the remainder is 0, we take the last number in the cycle of 9, which is 1.

So, the number in the unit’s place of 999^22292582 is 1.