Prove that a square of an integer ends with a 0, 1, 4, 5 6 or 9. (Hint: let n = 10k+c, where c = 0, 1, …,9)
To prove that a square of an integer ends with 0, 1, 4, 5, 6, or 9, we can use the hint provided. Let's consider an integer, n, and express it as n = 10k + c, where k is an integer and c can take on values between 0 and 9.
Now, let's square n:
n^2 = (10k + c)^2
Expanding this expression:
n^2 = 100k^2 + 20ck + c^2
We can observe that the terms 100k^2 and 20ck are both divisible by 10, which means they will always contribute a factor of 10 to the overall value of n^2. Therefore, we can focus on the last term, c^2, to determine the ones digit of n^2.
Let's consider the possible values of c:
1. c = 0:
In this case, c^2 = 0, and the ones digit of n^2 will be 0.
2. c = 1:
In this case, c^2 = 1, and the ones digit of n^2 will be 1.
3. c = 2:
In this case, c^2 = 4, and the ones digit of n^2 will be 4.
4. c = 3:
In this case, c^2 = 9, and the ones digit of n^2 will be 9.
5. c = 4:
In this case, c^2 = 16, and the ones digit of n^2 will be 6.
6. c = 5:
In this case, c^2 = 25, and the ones digit of n^2 will be 5.
7. c = 6:
In this case, c^2 = 36, and the ones digit of n^2 will be 6.
8. c = 7:
In this case, c^2 = 49, and the ones digit of n^2 will be 9.
9. c = 8:
In this case, c^2 = 64, and the ones digit of n^2 will be 4.
10. c = 9:
In this case, c^2 = 81, and the ones digit of n^2 will be 1.
From the above analysis, we can conclude that for any integer n, the square of n will always end with a digit of 0, 1, 4, 5, 6, or 9.