Question
Download Solution PDFFind the unit digit of (432)412 × (499)431.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
(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
∴ The unit digit of (432)412 × (499)431 is 4.
Last updated on Jul 5, 2025
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