What can be the last digit of a square of a natural number?

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:

Thus, the last digit of the square of a natural number can be 0, 1, 4, 5, 6, or 9.

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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.

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)