`12^(55)//3^(11)+8^(48)//16^(18)` will give the digit at units place as

Options

A. 4
B. 6
C. 8
D. 0

Answer

To find the unit digit of the expression \( \frac{12^{55}}{3^{11}} + \frac{8^{48}}{16^{18}} \), we will simplify each term step by step and then determine the unit digit.

### Step 1: Simplify the first term \( \frac{12^{55}}{3^{11}} \)

First, we can express 12 in terms of its prime factors:
\[ 12 = 3 \times 4 = 3 \times 2^2 \]
Thus,
\[ 12^{55} = (3 \times 2^2)^{55} = 3^{55} \times 2^{110} \]
Now substituting this back into the expression:
\[ \frac{12^{55}}{3^{11}} = \frac{3^{55} \times 2^{110}}{3^{11}} = 3^{55 - 11} \times 2^{110} = 3^{44} \times 2^{110} \]

### Step 2: Simplify the second term \( \frac{8^{48}}{16^{18}} \)

Next, we express 8 and 16 in terms of their prime factors:
\[ 8 = 2^3 \quad \text{and} \quad 16 = 2^4 \]
Thus,
\[ 8^{48} = (2^3)^{48} = 2^{144} \]
And,
\[ 16^{18} = (2^4)^{18} = 2^{72} \]
Now substituting this back into the expression:
\[ \frac{8^{48}}{16^{18}} = \frac{2^{144}}{2^{72}} = 2^{144 - 72} = 2^{72} \]

### Step 3: Combine the two simplified terms

Now we have:
\[ 3^{44} \times 2^{110} + 2^{72} \]

### Step 4: Find the unit digit of each term

**Finding the unit digit of \( 3^{44} \times 2^{110} \)**:
- The unit digit of powers of 3 cycles every 4: \(3, 9, 7, 1\).
- \(44 \mod 4 = 0\) implies the unit digit of \(3^{44}\) is \(1\).

- The unit digit of powers of 2 cycles every 4: \(2, 4, 8, 6\).
- \(110 \mod 4 = 2\) implies the unit digit of \(2^{110}\) is \(4\).

Now, the unit digit of \(3^{44} \times 2^{110}\) is:
\[ 1 \times 4 = 4 \]

**Finding the unit digit of \(2^{72}\)**:
- \(72 \mod 4 = 0\) implies the unit digit of \(2^{72}\) is \(6\).

### Step 5: Add the unit digits

Now we add the unit digits:
\[ 4 + 6 = 10 \]
The unit digit of \(10\) is \(0\).

### Final Answer

Thus, the digit at the unit place of the expression \( \frac{12^{55}}{3^{11}} + \frac{8^{48}}{16^{18}} \) is **0**.

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