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This essential trains for: SAT-I, GMAT, AMC-8, AMC-10.

Last digit of perfect squares

If a positive integer asdf3 ends in 3 what will its square end in?

Write the number in expanded form:

and square it:

Notice how, due to the final zero, neither the square of the larger part, nor the cross-term contribute to the last digit, since they both end in zero by construction.

Therefore, the last digit is given by the last digit of the square of the last digit.

For all the digits a positive integer can end with, the last digit of the square of the number is:

Last digit of n | Last digit of n2 |

0 | 0 |

1 | 1 |

2 | 4 |

3 | 9 |

4 | 6 |

5 | 5 |

6 | 6 |

7 | 9 |

8 | 4 |

9 | 1 |

We see that perfect squares can never end in 2, 3, 7 or 8.

Last digit of perfect cubes

Similarly, the last digit of perfect cubes is given by the table:

Last digit of n | Last digit of n3 |

0 | 0 |

1 | 1 |

2 | 8 |

3 | 7 |

4 | 4 |

5 | 5 |

6 | 6 |

7 | 3 |

8 | 2 |

9 | 9 |

We see that perfect cubes may end in any digit.

Not so with fourth powers which have very few options for their last digit:

Last digit of n | Last digit of n4 |

0 | 0 |

1 | 1 |

2 | 6 |

3 | 1 |

4 | 6 |

5 | 5 |

6 | 6 |

7 | 1 |

8 | 6 |

9 | 1 |

The successive positive integer powers of a number end digit forms repeating sequences as follows:

n ends in | n2 | n3 | n4 | n5 |

0 | 0 | 0 | 0 | 0 |

1 | 1 | 1 | 1 | 1 |

2 | 4 | 8 | 6 | 4 |

3 | 9 | 7 | 1 | 9 |

4 | 6 | 4 | 6 | 4 |

5 | 5 | 5 | 5 | 5 |

6 | 6 | 6 | 6 | 6 |

7 | 9 | 3 | 1 | 7 |

8 | 4 | 2 | 6 | 4 |

9 | 1 | 9 | 1 | 9 |

Example: What is the last digit of the sum:

First, we notice that the second term always ends in 6 regardless of the power.

Then we see that the powers of numbers that end in 7 end in digits that repeat every 4 powers. We need to find the remainder of dividing 401 by 4:

In the sequence of last digits, we need to position ourselves in the first position after a group of 4. Therefore, the last digit for the first term is 7 and the overall last digit is 3: