Find the **unit digit** of $$x = (164)^{169} + (333)^{337} - (727)^{726}$$.

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**Solution:**

1. We need to find the unit digit of:
   $$x = (164)^{169} + (333)^{337} - (727)^{726}$$

2. The unit digit of a number raised to a power depends only on the unit digit of the base.

3. Extract the unit digits of the bases:
   - Unit digit of 164 is 4
   - Unit digit of 333 is 3
   - Unit digit of 727 is 7

4. Find the unit digit of each term separately:

**Term 1:** $(164)^{169}$
   - Unit digit base = 4
   - Powers of 4 cycle every 2:
     - $4^1 = 4$ (unit digit 4)
     - $4^2 = 16$ (unit digit 6)
     - $4^3 = 4$ (unit digit 4)
     - $4^4 = 6$ (unit digit 6)
   - Since 169 is odd, unit digit = 4

**Term 2:** $(333)^{337}$
   - Unit digit base = 3
   - Powers of 3 cycle every 4:
     - $3^1 = 3$
     - $3^2 = 9$
     - $3^3 = 27$ (unit digit 7)
     - $3^4 = 81$ (unit digit 1)
   - Find remainder when 337 is divided by 4:
     $$337 mod 4 = 1$$
   - So unit digit = unit digit of $3^1 = 3$

**Term 3:** $(727)^{726}$
   - Unit digit base = 7
   - Powers of 7 cycle every 4:
     - $7^1 = 7$
     - $7^2 = 49$ (unit digit 9)
     - $7^3 = 343$ (unit digit 3)
     - $7^4 = 2401$ (unit digit 1)
   - Find remainder when 726 is divided by 4:
     $$726 mod 4 = 2$$
   - So unit digit = unit digit of $7^2 = 9$

5. Now substitute the unit digits:
   $$x = 4 + 3 - 9$$

6. Calculate the unit digit of the result:
   - $4 + 3 = 7$
   - $7 - 9 = -2$

7. Since unit digits are considered modulo 10, negative unit digit $-2$ is equivalent to $10 - 2 = 8$.

**Answer:** $8$

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