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Given:

(432)412 × (499)431

Concept:

9even no. = unit digit 1

9odd 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,

 Power of 2 Last digit 21 2 22 4 23 8 24 6 25 2 26 4 27 8 28 6 29 2

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

Quantitative Value Definition

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