Is the following claim true, false or sometimes true?

Hi there,

here's my question:

*When you square any integer, the last digit of the number you get will never be an eight.*

Is this statement true, false, or sometimes true. Justify your result.

so..

2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2 = 100

as you can see this is true for single digits i've tried double digits as well up till 30^2, and clearly enough the last digit is never an eight.

from what i've got above im guessing that this statement is true!

Is there another way to justify this instead of just showing squares of random numbers? maybe some sort of equation? or a better way to prove this statement?

Re: Is the following claim true, false or sometimes true?

Quote:

Originally Posted by

**drake01** 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2=36, 7^2=49, 8^2=64, 9^2=81, 10^2 = 100

You forgot 1^2 = 1.

Represent an integer k as 10m + n for some integers m, n such that 0 <= n < 9. (Note that this also works when k < 0.) Then consider k^2.

Re: Is the following claim true, false or sometimes true?

You might be correct, but you haven't shown it.

Maybe try a proof by induction?

What about ending in 2,3 or 7 as well? Have you found any of those?

Re: Is the following claim true, false or sometimes true?

every integer can be written as

$\displaystyle a(10)^n + b(10)^{n-1} + c(10)^{n-2} + ... + d$ , where a , b , c, ... are integral constants, and the last term, d , is the units digit of that integer.

the square of this general integer ...

$\displaystyle [a(10)^n + b(10)^{n-1} + c(10)^{n-2} + ... + d]^2$ = the sum of many terms that have at least one factor of a power of 10 + the last term $\displaystyle d^2$

since d is one of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , the units digit of $\displaystyle d^2$ belongs to the set {0, 1, 4, 5, 6, 9} ... the digit 8 is not a member of that set, so the statement is true.