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Digits of a Square

Date: 05/26/2001 at 12:14:38
From: Jeevanjyoti Chakraborty
Subject: Number Theory

If the tens digit of a^2 (a is an integer) is 7, what is the units 

  My attempt:  a^2 = N*100+10*7+x
               x can be 0,1,4,5,6,9

After this I could not proceed.

Date: 05/26/2001 at 18:50:49
From: Doctor Pete
Subject: Re: Number Theory


Consider numbers of the form 50k+m, where k and m are nonnegative 
integers, and m < 50.  We have

     (50k+m)^2 = 2500k^2 + 100km + m^2.

This is congruent to m^2 modulo 100, since the first two terms are 
divisible by 100 for all values of m and k. In other words, the tens 
and units digits of (50k+m)^2 are the same as those for m^2. This 
means we only need to consider squares less than 50^2. At this point, 
we can simply compute the list of squares and find the answer, but we 
can do more analysis. If we consider numbers of the form 25k+m, this 
time m < 25, we find

     (25k+m)^2 - (25k-m)^2 = ((25k+m)+(25k-m))((25k+m)-(25k-m))
                           = (50k)(2m)
                           = 100km,

which is congruent to 0 modulo 100; i.e., this number is divisible 
by 100. It follows that the difference of the squares (25k+m)^2 and 
(25k-m)^2 is a number with tens and units digits equal to 0; hence the 
square (25k+m)^2 has the same last two digits as the square (25k-m)^2.  
So now we only need to consider squares less than 25^2. From here the 
simplest way to solve the problem is to compute the first 25 squares; 
we find that the only square with tens place equal to 7 is 576, which 
is the square of 24.  Therefore, the only possible value of the units 
digit is 6.

- Doctor Pete, The Math Forum 

Date: 07/16/2003 at 22:36:04
From: Anonymous
Subject: Perfect square numbers

This question is from the 1999 Mathematics Competition papers out 

"If the tens digit of a perfect square number is 7, how many units 
digits are possible?"

a) one   b) two   c) three   d) four   e) five

The answer to the above question is a) one .


Date: 07/17/2003 at 12:36:56
From: Doctor Peterson
Subject: Re: Perfect square numbers


The first thing I would do here would be to "play" with the problem, 
squaring numbers starting at 1 and seeing what the last two digits 
are, looking for a pattern. I know that the last two digits of a 
square depend only on the last two digits of the number being squared 
(do you see why?), so I wouldn't have to go past 100 to get all 
possibilities; but I'd hope to find a shortcut.

Rather than do that, see Doctor Pete's answer, above. As you can see, 
he is basically just reducing the number of cases to test by making 
use of some symmetries. If I had gone past 25 in my "play," I would 
have run up against this fact, that the same set of 25 final-digit-
pairs repeats four times in this pattern:

  0          25          50          75          100
   --------->  <---------  --------->  <---------

Note that another way to prove the (25k+m) fact he came up with is to 
express it this way:

  (50-m)^2 = 2500m^2 - 100m + m^2 = 100(25m^2 - m) + m^2

is congruent to m^2 modulo 100.

- Doctor Peterson, The Math Forum 

Date: 07/17/2003 at 13:14:03
From: Doctor Greenie
Subject: Re: Perfect square numbers

Hello -

Dr. Peterson has shown you an analysis of this problem that uses a 
sophisticated "trick" of noting that the squares of 50x+y and 50x-y 
(x and y integers) have the same final two digits.

Here is an analysis that is more straightforward.  Being more 
straightforward, it is an approach which might more reasonably be 
expected to be used in a contest situation; however, being less 
sophisticated, it probably involves a bit more work....

The last two digits of the square of a number are determined by the 
last two digits of the number.  To see this, let a and b be integers 
and consider the two integers a and 100a+b.  These two integers have 
the same last two digits.  And their squares also have the same last 
two digits:

  (100a+b)^2 = 10000a^2 + 200(ab) + b^2 = 100(100a^2 + 2b) + b^2

So let's look at all two-digit numbers to see which ones have squares 
with tens digit 7.  We now let x and y be single digits, and we let 
our two-digit number be


When we square this two-digit integer, we find

  (10x+y)^2 = 100x^2 + 20xy + y^2

We want to find when this square number can have tens digit 7. The 
100x^2 does not contribute anything to the tens digit. The tens digit 
of this square number is the units digit of the integer part of 
20xy + y^2 divided by 10, which we can write as

  [(20xy + y^2)/10] mod 10


  (2xy + [y^2/10]) mod 10

Now we simply try all the different possible digits for y and see 
which ones give us

  (2xy + [y^2/10]) mod 10 = 7 

   y      equation        possible value(s) of x
   0   0x + 0 mod 10 = 7  (none)
   1   2x + 0 mod 10 = 7  (none)
   2   4x + 0 mod 10 = 7  (none)
   3   6x + 0 mod 10 = 7  (none)
   4   8x + 1 mod 10 = 7  x=2; x=7
   5  10x + 2 mod 10 = 7  (none)
   6  12x + 3 mod 10 = 7  x=2; x=7
   7  14x + 4 mod 10 = 7  (none)
   8  16x + 6 mod 10 = 7  (none)
   9  18x + 8 mod 10 = 7  (none)

The only 2-digit perfect square numbers with tens digit 7 are 24, 26, 
74, and 76. For each of these, the units digit of the square is 6, so 
the answer to the question is "one."

I hope this helps further.

- Doctor Greenie, The Math Forum 
Associated Topics:
High School Number Theory

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