The user-friendly version of this content is available here.

The following content is copyright (c) 2009-2013 by Goods of the Mind, LLC.

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:

equation

and square it:

equation

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 nLast digit of n2
00
11
24
39
46
55
66
79
84
91

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 nLast digit of n3
00
11
28
37
44
55
66
73
82
99

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 nLast digit of n4
00
11
26
31
46
55
66
71
86
91

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:

equation

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:

equation

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:

equation