The digit in the unit's place of `[(251)^(98)+(21)^(29)-(106)^(100)+(705)^(35)-16^(4)+259]` is :

Options

A. 1
B. 4
C. 5
D. 6

Answer

To find the digit in the unit's place of the expression \((251)^{98} + (21)^{29} - (106)^{100} + (705)^{35} - (16)^{4} + 259\), we will analyze each term separately and focus only on the unit's digit of each term.

### Step 1: Find the unit's digit of \( (251)^{98} \)

The unit's digit of \( 251 \) is \( 1 \).
Any power of \( 1 \) will always have a unit's digit of \( 1 \).

**Unit's digit of \( (251)^{98} \) is \( 1 \).**

### Step 2: Find the unit's digit of \( (21)^{29} \)

The unit's digit of \( 21 \) is \( 1 \).
Similar to the previous case, any power of \( 1 \) will always have a unit's digit of \( 1 \).

**Unit's digit of \( (21)^{29} \) is \( 1 \).**

### Step 3: Find the unit's digit of \( (106)^{100} \)

The unit's digit of \( 106 \) is \( 6 \).
The unit's digits of powers of \( 6 \) are always \( 6 \) (i.e., \( 6^1 = 6, 6^2 = 36, 6^3 = 216, \ldots \)).

**Unit's digit of \( (106)^{100} \) is \( 6 \).**

### Step 4: Find the unit's digit of \( (705)^{35} \)

The unit's digit of \( 705 \) is \( 5 \).
The unit's digits of powers of \( 5 \) are always \( 5 \) (i.e., \( 5^1 = 5, 5^2 = 25, 5^3 = 125, \ldots \)).

**Unit's digit of \( (705)^{35} \) is \( 5 \).**

### Step 5: Find the unit's digit of \( (16)^{4} \)

The unit's digit of \( 16 \) is \( 6 \).
As established earlier, the unit's digits of powers of \( 6 \) are always \( 6 \).

**Unit's digit of \( (16)^{4} \) is \( 6 \).**

### Step 6: Find the unit's digit of \( 259 \)

The unit's digit of \( 259 \) is \( 9 \).

**Unit's digit of \( 259 \) is \( 9 \).**

### Step 7: Combine all unit's digits

Now we combine the unit's digits we found:
- From \( (251)^{98} \): \( 1 \)
- From \( (21)^{29} \): \( 1 \)
- From \( (106)^{100} \): \( 6 \)
- From \( (705)^{35} \): \( 5 \)
- From \( (16)^{4} \): \( 6 \)
- From \( 259 \): \( 9 \)

Putting it all together:
\[ 1 + 1 - 6 + 5 - 6 + 9 \]

Calculating step by step:
1. \( 1 + 1 = 2 \)
2. \( 2 - 6 = -4 \) (which has a unit's digit of \( 6 \) when considering modulo \( 10 \))
3. \( 6 + 5 = 11 \) (unit's digit \( 1 \))
4. \( 1 - 6 = -5 \) (which has a unit's digit of \( 5 \))
5. \( 5 + 9 = 14 \) (unit's digit \( 4 \))

Thus, the final unit's digit of the entire expression is \( 4 \).

### Final Answer:
The digit in the unit's place of the expression is \( 4 \).

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