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**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

⇒ 2^{412} × 9^{431}

As we know unit digit of 2^{1} = 2, 2^{2} = 4, 2^{3} = 8, 2^{4} = 6

⇒ 2^{4(103) }× 9^{431}

⇒ 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 |

## Quantitative Value Pdf Free Download Free

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.