The square of an integer never hasa last digit equal to:a.1 b.4 c.5 d.8 e.9
a number ending in 0 when squared ends in 0
a number ending in 1 when squared ends in 1
a number ending in 2 when squared ends in 4
a number ending in 3 when squared ends in 9
...
a number ending in 9 when squared ends in 1
investigate.
http://www.mathwarehouse.com/arithmetic/numbers/list-of-perfect-squares.php
To determine which last digits are not possible when squaring an integer, let's analyze the last digit of the numbers 0 to 9 after squaring them:
0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
From the above calculations, we can observe that the last digits can be 0, 1, 4, 5, 6, or 9.
Therefore, the last digit that is not possible when squaring an integer is 8 (d), as it does not appear in the list of possible last digits.
To determine the answer to this question, we need to analyze the patterns in the last digit of the squares of integers.
Here's how you can find the answer:
1. Start by listing the last digit of the squares of the integers.
- For example, if we square integers from 0 to 9, we get:
- 0^2 = 0
- 1^2 = 1
- 2^2 = 4
- 3^2 = 9
- 4^2 = 16
- 5^2 = 25
- 6^2 = 36
- 7^2 = 49
- 8^2 = 64
- 9^2 = 81
2. Analyze the last digits and see if there's any repeating pattern.
- By examining the last digits, we can observe that the squares of 0 and 1 have the last digits 0 and 1, respectively.
- The squares of 2 and 8 end in the last digit 4.
- The squares of 3 and 7 have the last digit 9.
- The squares of 4 and 6 end in the last digit 6.
- The squares of 5 and 9 have the last digit 5.
3. Determine which options are never the last digit of a square.
- Comparing the observed patterns with the answer choices, we can see that option (c) and option (d) are never the last digit of a square.
- Therefore, the answer is (c) and (d), which are 5 and 8.
So, the square of an integer never has a last digit equal to 5 or 8.