Which are the digits that never appear in any units' digit of any square number?

He's a maths problem I am stuck with... Thanks for your help in advance.

List the digits that will **not** be in the units' digit of any square number? (please show clear reasoning)

Re: Which are the digits that never appear in any units' digit of any square number?

Quote:

Originally Posted by

**Natasha1** He's a maths problem I am stuck with... Thanks for your help in advance.

List the digits that will **not** be in the units' digit of any square number? (please show clear reasoning)

Hi Natasha1

When you multiply 2 numbers together you can always tell what the last digit is going to be just by looking at the last digits of the initial 2 numbers.

eg $\displaystyle 3129\times123750856 \text{ will end in a 4 because } 9\times 6=54 \text { and 54 ends in a 4 } $

with this knowledge you can work out what are the possible last digits of squared numbers can be and hence, what digits they cannot be.

Re: Which are the digits that never appear in any units' digit of any square number?

So:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100...

the digits are 1, 4, 9, 6, 5 and 0 so the ones it can't be are 2, 3, 7 and 8.

But how can I prove it?

Re: Which are the digits that never appear in any units' digit of any square number?

Quote:

Originally Posted by

**Natasha1** So:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100...

the digits are 1, 4, 9, 6, 5 and 0 so the ones it can't be are 2, 3, 7 and 8.

But how can I prove it?

Note that for numbers larger than 10 that you are simply repeating the list...11^2 will have a 1 in the units position because 1 is in the accepted list. Likewise 13^2 will end in a 9. etc.

-Dan

Re: Which are the digits that never appear in any units' digit of any square number?

Well, proofs are definitely not my forte but I can at least give you more of an explanation

$\displaystyle \begin{align*}\text{Let y }&=(10a+x) \text{ where }a, x, \text{ and therefore y are in the set of integers greater or equal to zero, also } x<10\\y^2&=100a^2+20ax+x^2\\y^2&=10(10a^2+2ax)+x^2 \end{align*}$

so, the last digit will be the last digit of $\displaystyle x^2$

OR

If you think about how long multplication works you can also 'see' that this will be true.

Is this enough or are you after a formal proof?