To determine which of the given numbers [tex]\(123^2, 77^2, 82^2, 161^2,\)[/tex] or [tex]\(109^2\)[/tex] ends with the digit 1, we will consider the properties of the last digits of the numbers and the resulting last digits of their squares.
1. 123:
- Last digit of 123 is 3.
- The square of a number ending in 3 (i.e., [tex]\(3^2\)[/tex]) ends with 9 because [tex]\(3 \cdot 3 = 9\)[/tex].
- Thus, [tex]\(123^2\)[/tex] ends with 9.
2. 77:
- Last digit of 77 is 7.
- The square of a number ending in 7 (i.e., [tex]\(7^2\)[/tex]) ends with 9 because [tex]\(7 \cdot 7 = 49\)[/tex].
- Thus, [tex]\(77^2\)[/tex] ends with 9.
3. 82:
- Last digit of 82 is 2.
- The square of a number ending in 2 (i.e., [tex]\(2^2\)[/tex]) ends with 4 because [tex]\(2 \cdot 2 = 4\)[/tex].
- Thus, [tex]\(82^2\)[/tex] ends with 4.
4. 161:
- Last digit of 161 is 1.
- The square of a number ending in 1 (i.e., [tex]\(1^2\)[/tex]) ends with 1 because [tex]\(1 \cdot 1 = 1\)[/tex].
- Thus, [tex]\(161^2\)[/tex] ends with 1.
5. 109:
- Last digit of 109 is 9.
- The square of a number ending in 9 (i.e., [tex]\(9^2\)[/tex]) ends with 1 because [tex]\(9 \cdot 9 = 81\)[/tex].
- Thus, [tex]\(109^2\)[/tex] ends with 1.
Clearly, the numbers whose squares end with the digit 1 are 161 and 109.
So, [tex]\(161^2\)[/tex] and [tex]\(109^2\)[/tex] end with the digit 1.
