|
Question 1207680: The numbers $24^2 = 576$ and $56^2 = 3136$ are examples of perfect squares that have a units digits of $6.$
If the units digit of a perfect square is $5,$ then what are the possible values of the tens digit?
Found 3 solutions by greenestamps, math_tutor2020, Edwin McCravy:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
ANSWER: 2 only
If the perfect square n^2 has units digit 5, then n itself has units digit 5, so it is of the form 10m+5, where m is an integer. Then
So n^2 is equal to 100 times some integer, plus 25; that means the last two digits are always 25.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 2 is the only possibility.
Explanation
Let's square the first few positive integers.
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
10^2 = 100
11^2 = 121
Once reaching 11, the units digit rule will follow the same idea as 1^2 = 1.
Meaning that we have encapsulated all possible units digits in that list above.
The possible units digits are: 0, 1, 4, 5, 6, 9
The only time 5 shows up as a units digit is when 5^2 = 25
Therefore x^2 will have a units digit of 5 if and only if x ends with 5.
Examples
5^2 = 25
15^2 = 225
25^2 = 625
35^2 = 1225
45^2 = 2025
Not only does each result end in 5, it also ends with 25.
For proof of this, refer to the solution written by tutor greenestamps.
Notice how m = 0 is the smallest possible integer value of m such that 10m+5 is positive.
So the smallest 100(m^2+m) can get is 100(0^2+0) = 0 which won't affect the +25 at the end.
If we tried m = 1 then 100(m^2+m) = 100(1^2+1) = 200; but notice the trailing zeros. Specifically the trailing zero in the tens digit. Adding this particular zero wont change the 2 in 25 at the end.
As m gets larger, so does 100(m^2+m) which contributes more trailing zeros. This is why the +25 at the end will stick around in each situation.
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website!
It's also true that the square of any positive integer that ends with 5 either
ends with 025, 225, or 625. (It's true for the square of 5 if you consider
25 as 025). If you want a proof of this, just ask.
Edwin
|
|
|
| |
