What is the unit place number in $(2137)^753$ ?

Q: What is the unit place number in $(2137)^753$ ?

To determine the unit place number of $ (2137)^{753} $, we need to focus on the unit digit of the base number, which is 7. The unit digit of a number raised to any power depends only on the unit digit of the base number. Therefore, we will examine the pattern of the unit digits of powers of 7. Let's calculate the unit digits of the first few powers of 7: 1. $ 7^1 = 7 $ (unit digit is 7) 2. $ 7^2 = 49 $ (unit digit is 9) 3. $ 7^3 = 343 $ (unit digit is 3) 4. $ 7^4 = 2401 $ (unit digit is 1) 5. $ 7^5 = 16807 $ (unit digit is 7) We observe that the unit digits repeat every four powers: 7, 9, 3, 1. This is a repeating cycle of length 4. To find the unit digit of $ (2137)^{753} $, we need to determine the position of 753 in this cycle. We do this by finding the remainder when 753 is divided by 4: \[ 753 \div 4 = 188 \text{ remainder } 1 \] The remainder is 1, which means $ 753 \equiv 1 \pmod{4} $. Therefore, the unit digit of $ (2137)^{753} $ corresponds to the unit digit of $ 7^1 $, which is 7. Thus, the unit place number in $ (2137)^{753} $ is 7. Now, let's explain why each of the other options is incorrect: 1. **Option 1 (1):** The unit digit 1 appears in the cycle at $ 7^4 $, $ 7^8 $, etc. Since 753 leaves a remainder of 1 when divided by 4, it does not correspond to this position in the cycle. 2. **Option 2 (3):** The unit digit 3 appears in the cycle at $ 7^3 $, $ 7^7 $, etc. Since 753 leaves a remainder of 1 when divided by 4, it does not correspond to this position in the cycle. 3. **Option 4 (9):** The unit digit 9 appears in the cycle at $ 7^2 $, $ 7^6 $, etc. Since 753 leaves a remainder of 1 when divided by 4, it does not correspond to this position in the cycle. Therefore, the correct answer is Option 3, which is 7.

Question

Accepted Answer

To determine the unit place number of $ (2137)^{753} $, we need to focus on the unit digit of the base number, which is 7. The unit digit of a number raised to any power depends only on the unit digit of the base number. Therefore, we will examine the pattern of the unit digits of powers of 7.

Let's calculate the unit digits of the first few powers of 7:

1. $ 7^1 = 7 $ (unit digit is 7)
2. $ 7^2 = 49 $ (unit digit is 9)
3. $ 7^3 = 343 $ (unit digit is 3)
4. $ 7^4 = 2401 $ (unit digit is 1)
5. $ 7^5 = 16807 $ (unit digit is 7)

We observe that the unit digits repeat every four powers: 7, 9, 3, 1. This is a repeating cycle of length 4.

To find the unit digit of $ (2137)^{753} $, we need to determine the position of 753 in this cycle. We do this by finding the remainder when 753 is divided by 4:

$$ 753 \div 4 = 188 \text{ remainder } 1 $$

The remainder is 1, which means $ 753 \equiv 1 \pmod{4} $. Therefore, the unit digit of $ (2137)^{753} $ corresponds to the unit digit of $ 7^1 $, which is 7.

Thus, the unit place number in $ (2137)^{753} $ is 7.

Now, let's explain why each of the other options is incorrect:

1. **Option 1 (1):** The unit digit 1 appears in the cycle at $ 7^4 $, $ 7^8 $, etc. Since 753 leaves a remainder of 1 when divided by 4, it does not correspond to this position in the cycle.

2. **Option 2 (3):** The unit digit 3 appears in the cycle at $ 7^3 $, $ 7^7 $, etc. Since 753 leaves a remainder of 1 when divided by 4, it does not correspond to this position in the cycle.

3. **Option 4 (9):** The unit digit 9 appears in the cycle at $ 7^2 $, $ 7^6 $, etc. Since 753 leaves a remainder of 1 when divided by 4, it does not correspond to this position in the cycle.

Therefore, the correct answer is Option 3, which is 7.