Respuesta :
Answer:
Any of {0, 1, 4, 5, 6, 9}
Step-by-step explanation:
The squares of the digits 0–9 are ...
0, 1, 4, 9, 16, 25, 36, 49, 64, 91
The last digit of a square must be the last digit of one of these:
... {0, 1, 4, 5, 6, 9}
_____
Comment on the problem
You will see that the squares of 5 and 6 end in 5 and 6, respectively. Such a number is called an automorph. The last half of its square is the original number. These can be extended indefinitely, usually according to a simple rule. The 3-digit automorph ending in 6 is 376. (376² = 141,376). The 5-digit automorph ending in 5 is 90625. (90625² = 8212890625)
Automorphs can also be found in number bases other than base-10.
The last digit of the square of a natural number can be 0, 1, 4, 5, 6, or 9.
A natural number is a number resulting from a count of a finite number of things. Thus, integers from 0 are considered natural numbers, since they allow us to represent a certain amount of things.
To determine what can be the last digit of a square of a natural number, this number must be multiplied by itself:
- 1 x 1 = 1
- 2 x 2 = 4
- 3 x 3 = 9
- 4 x 4 = 16
- 5 x 5 = 25
- 6 x 6 = 36
- 7 x 7 = 49
- 8 x 8 = 64
- 9 x 9 = 81
- 10 x 10 = 100
Thus, the last digit of the square of a natural number can be 0, 1, 4, 5, 6, or 9.
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