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

1. ## 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)

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

3. ## 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?

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

5. ## 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?

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