SOLUTION: 𝑘 is an integer. The last two digits of 𝑘^2 are equal to each other. What does this tell you about the last digit of 𝑘^2? What does it tell you about 𝑘?

Algebra ->  Finance -> SOLUTION: 𝑘 is an integer. The last two digits of 𝑘^2 are equal to each other. What does this tell you about the last digit of 𝑘^2? What does it tell you about 𝑘?       Log On


   

Question 1204004: 𝑘 is an integer. The last two digits of 𝑘^2 are equal to each other.
What does this tell you about the last digit of 𝑘^2?
What does it tell you about 𝑘?
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let the last 2 digits of k be x and y, so that the number is 10x+y. Then k^2 is 100x^2+20xy+y^2.

We want to know the condition(s) that make the last two digits of k^2 = 100x^2+20xy+y^2 the same.

The last 2 digits of 100x^2 are always "00", so the last 2 digits of k^2 are determined by 20xy+y^2.

So, for each last digit y, we want to find the values of next-to-last digit x for which the last 2 digits of 20xy+y^2 are the same.

  last digit y  20xy+y^2  value(s) of x that make last 2 digits of 20xy+y^2 the same
 ---------------------------------------------------------
      0            0        any (trivial case -- the square of any integer with last digit 0 has last 2 digits 00)
      1         20x+1       none
      2         40x+4       1 or 6 (last 2 digits "44")
      3         60x+9       none
      4         80x+16      none
      5        100x+25      none
      6        120x+36      none
      7        140x+49      none
      8        160x+64      3 or 8 (last 2 digits "44")
      9        180x+81      none

Summary....

The last 2 digits of k^2 are the same if, and only if...
(1) the last digit of k is 0 (last 2 digits of k^2 "00"); or
(2) the last 2 digits of k are 12 or 62, or 38 or 88 (last 2 digits of k^2 "44")

It's not clear what kind of answers are wanted for the questions that are asked, so I leave it to you to use those results to answer them.

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Let m = the number formed by the last two (like) digits only.
Let n be the number formed by k^2 with the last two digits chopped off.
Then k^2 = 100n+m, where m = 00,11,22,33,44,55,66,77,88, or 99

Since 100n is divisible by 4, then the remainder when k^2 is divided by 4
is the same as the remainder when m is divided by 4.

k^2 must end with same last digit as the last digit of the square of the last
digit of k.

0^2=0, 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, 9^2=81

So we can rule out m as 22,33,77,88

So m is one of these:  00,11,44,55,66,99

If m and k^2 are even, then m is one of these: 00, 44, or 66.  But an even
number is divisible by 4, but 66 isn't divisible by 4, so 66 is ruled out as a
possibility for m.

That leaves only 00,11,44,55,99

If m and k^2 are odd, then m is 11, 55 or 99.  Since k^2 is odd, k is also odd
and k = 2q+1 for some nonnegative integer q.

Then k^2 = (2q+1)^2 = 4q^2+4q+1 = 4(q^2+q)+1 

so when k^2 or m are divided by 4, the remainder must be 1. However, when 11 is
divided by 4, the quotient is 2 and the remainder is 3.  When 55 is divided by
4, the quotient is 13 and the remainder is also 3. When 99 is divided by 4, the
quotient is 24 and the remainder is also 3.  But the remainder must be 1, not 3.

So 11,33,99 are ruled out.

Therefore, m = 00 or 44.  So the last digit of k^2 is 0 or 4 and so the last
digit of k is 0, 2, or 8.

Edwin