Find the unit digit of (432)⁴¹² × (499)⁴³¹.

Given:

(432)⁴¹² × (499)⁴³¹

Concept:

9^{even no.} = unit digit 1

9^{odd no.} = unit digit 9

Calculation:

(432)⁴¹² × (499)⁴³¹

Taking unit digits

⇒ 2⁴¹² × 9⁴³¹

As we know unit digit of 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 6

⇒ 2⁴⁽¹⁰³⁾ × 9⁴³¹

⇒ 6 × 9

⇒ 54

∴ The unit digit of (432)⁴¹² × (499)⁴³¹ is 4.

To determine the last digit of the number 432412, we need to focus on the last digit of base 432 i.e. 2 and the exponential part 412.

We know,

Notice the pattern of the last digit. It is 2, 4, 8, 6, 2, 4, 8, 6, 2 …… so on.

Thus the last digit is repetitive and is a four-digit long i.e. 1, 2, 8, 6. If we keep on writing this table till the power of 2 reaches 412 then how many times this pattern repeated can be found by dividing 412 by 4.

412 divided by 4 is 103 with remainder 0 which indicates that the pattern gets fully repeated 412 times and then ends up with the digit i.e. 4. (if it is fully divisible we take power as 4)

∴ The Last digit of the number 432412 is 6.

9even no. = unit digit 1

9odd no. = unit digit 9

∴ The Last digit of the number 9431 is 9

∴ The unit digit of (432)412 × (499)431 is 4.