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A very interesting math problem! ! !
Properties of 1) Complete Square Number

If a number is the complete square of another integer, then we call it the complete square number, also called the square number. For example:

0, 1,4,9, 16,25,36,49,64,8 1, 100, 12 1, 144, 169, 196,225,256,289,324,36 1,400,44 1,484,…

By observing these complete square numbers, we can understand the regularity of their single digits, ten digits and the sum of numbers. Let's study some common properties of complete squares:

Attribute 1: The last digit of a complete square number can only be 0, 1, 4, 5, 6, 9.

Property 2: The single digit of the square of an odd number is odd, and the ten digits are even.

Prove that odd numbers must be one of the following five forms:

10a+ 1, 10a+3, 10a+5, 10a+7, 10a+9

After squaring separately, we get

( 10a+ 1)= 100+20a+ 1 = 20a(5a+ 1)+ 1

( 10a+3)= 100+60a+9 = 20a(5a+3)+9

( 10a+5)= 100+ 100 a+25 = 20(5a+5a+ 1)+5

( 10a+7)= 100+ 140 a+49 = 20(5a+7a+2)+9

( 10a+9)= 100+ 180 a+8 1 = 20(5a+9a+4)+ 1

To sum up, we can know that the square of odd numbers and the unit number are odd numbers 1, 5, 9; Ten digits are even.

Property 3: If the ten digits of a complete square number are odd, then its single digits must be 6; On the other hand, if the single digit of a complete square number is 6, then its ten digits must be odd.

Prove that it is known = 10k+6, and prove that k is odd. Because the unit number of is 6 and the unit number of m is 4 or 6, we can set m= 10n+4 or 10n+6. rule

10k+6 =( 10n+4)= 100+(8n+ 1)x 10+6

Or10k+6 = (10n+6) =100+(12n+3) x10+6.

That is k =10+8n+1= 2 (5+4n)+1.

Or k =10+12n+3 = 2 (5+6n)+3.

K is very strange.

Inference 1: If all ten digits of a number are odd numbers and one of them is not 6, then the number must not be completely square.

Inference 2: If the single digit of a complete square number is not 6, then its ten digits are even.

Property 4: the square of even number is a multiple of 4; The square of an odd number is a multiple of 4 plus 1.

This is because (2k+1) = 4k (k+1)+1.

(2k)=4

Property 5: the square of odd number is 8n+ 1 type; The square of even number is 8n or 8n+4 type.

In the proof of property 4, from k(k+ 1) being an even number, it can be concluded that (2k+ 1) is a number of type 8n+ 1; From the odd or even numbers, (2k) can be 8n or 8n+4.

Attribute 6: Square number must be in one of the following two forms: 3k, 3k+ 1.

Because natural numbers are divided by 3, they can be divided into three categories according to the difference of remainder: 3m, 3m+ 1, 3m+2. After squaring, you get.

(3m)=9=3k

(3m+ 1)= 9+6m+ 1 = 3k+ 1

(3m+2)=9+ 12m+4=3k+ 1

Similarly, you can get:

Property 7: The square of a number that is not divisible by 5 is 5k 1, and the square of a number that is divisible by 5 is 5k.

Attribute 8: The square number has one of the following forms: 16m,16m+,16m+4, 16m+9.

In addition to the above properties about single digits, ten digits and remainder, we can also study the sum of all digits of a complete square number. For example, the digits of 256 add up to 2+5+6 = 13, and 13 is the sum of digits of 256. If you add up the digits of 13: 1+3=4, 4 can also be called the sum of digits of 256. The sum of the digits of a number mentioned below refers to the addition of its digits. If the sum of the obtained numbers is not a number, the obtained numbers are added again until it becomes a number. We can get the following proposition:

The sum of a number is equal to the remainder of this number divided by 9.

Let's take four digits as an example to illustrate this proposition.

Let's make it four digits.

= 1000 a+ 100 b+ 10c+d

= 999a+99b+9c+(a+b+c+d)

= 9( 1 1 1a+ 1 1 b+ c)+(a+b+ c+d)

Obviously, a+b+c+d is the remainder of four digits divided by 9.

For n digits, you can also imitate this method to prove.