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# The units digit of a perfect square is 6. What are the possible values of the tens digit?

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The units digit of a perfect square is 6. What are the possible values of the tens digit?

Mellie  Jul 13, 2015

#1
+17744
+10

1, 3, 5, 7, 9

geno3141  Jul 13, 2015
#1
+17744
+10

1, 3, 5, 7, 9

geno3141  Jul 13, 2015
#2
+5

Could you explain how you got that?

Guest Jul 24, 2016
#3
+92805
0

Hi guest,

That is a great question.  Why don't you join up.  We want to get to know people like you. :)

Your questions will be given priority too if you are a member, plus they will be auto saved in your watch list so you won't lose track of them :)

Melody  Jul 24, 2016
#4
+92805
+1

My solution is not very elegant I am afraid but here goes.

When you multiply any 2 numbers together, the resultant last digit will be the last digit of the product of the original last digits

for instance.

536734 * 205446285297 must end in 8.

I did not do the multiplication.  I just looked at the last digits 4 and 7

4 * 7 = 28 the last digit is 8 so the last digit of the original product will also be 8.

Also, the resultant last TWO digits will be the the last 2 digits of the product of just the last two digits.

So

536734 * 205446285297 must end in in 98 because 34*97=3298, the last 2 digits are 98

If you set up a couple of long multiplications it should become clear to you why this is so.

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Now is a perfect square ends in 6, then the last digit squared must end in 6 also

What single digit numbers squared end in 6?    4*4=16   and   6*6=36   they are  the only 2 so the number that is being squared must end in 4 or 6

Let the second last digit be A.  A can be any digit.

So the last 2 digits of our number that is being squared are either   A6   or    A4

That is, the last 2 digits of the number are either (10A+6)   or   (10A+4)

Lets square these.

NOTE:

If the algebra is too muc for you, just look at all the possibilities one by one.

that is

$$06^2\quad 16^2 \quad 26^2 \quad 36^2 \quad 46^2 \quad 56^2 \quad 66^2 \quad up\;to\;96^2$$

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$$(10A+6)^2=100A^2+120A+36\\ \mbox{the 100A^2 will not effect the second digit so I can forget about that one}\\ 120A+36\\ 120A+30+6\\ 30(4A+1)+6\\ 3(4A+1)*10+6\\ \mbox{The second last digit will be the last digit of }3(4A+1)\\ 3(4A+3)=12A+9=10A+2A+9\\ \mbox{The 10A is not going to effect the digit we want so the 2nd last digit will be the last digit of 2A+9}\\ \mbox{This is any even number plus 9. The last digit could be any odd number}.$$

If the last digit is 4 then I can go through the same rigmarole for (10A+4)2

Anyway, this is why the second last digit must be an odd number :)

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Muck around with some long multiplications and this will become clear to you.

Questions like this are really good for developing a real understanding of how numbers work :)

Melody  Jul 24, 2016
#5
+87301
+1

Notice that a perfect square ending in 6 will be formed for a = 0,1,2,3.........

whenever we have either:

[10a + 4]^2   =   100a^2 +  80a + 16      ...... or.....

[10a + 6]^2  =   100a^2 + 120a + 36

The cyclic pattern for the 10s digit in the first expression is :

1, 9, 7, 5, 3...........

And the cyclic pattern for the 10s digit in the second expression is :

3, 5, 7, 9, 1...........

So......in this case, the 10s digit will always be odd

CPhill  Jul 24, 2016