A four digit number is the square of a certain number, and the first two digits and the last two dig

Answer: Assuming that the last two digits are not 0, I don't know if the last two digits are 0).

Analysis: Let this four-digit number be (10a+b) squared, i.e., 100a + 20ab + b, where a is 3 to 9 and b is 1 to 9 (when a is 3, b is 4 to 9).

where 20ab=100a+10b (0<=b<=9), where 4<=ab<=81 (a=4, b=1 is 4, a=9, b=9 is 81).

At this time, you can try one by one: when b = 1, b = 0 or 8 can be a square number, b = 0 when a has no solution, b = 8 when a = 4, and then substitute, 41 is in line.

When b = 2, b = 0 or 6 can be a square number, b = 0 when a = 5, b = 6 a = 4 or 9, and then substitute, 41 matches.

When a=5, a=11 can make the upper two digits 36, which is the square number, and 2ab must be less than 1100. In the same way, 5, so the high two digits must be the square of a. So 20ab<100 is ab<5

So only 41 matched.

Let a two-digit number be 10x+y, and its square = 100x 2+y(20x+y);

From 100x 2, we can know that x 4, and from y(20x+y) we know x 4, so if there is an answer to this question, x can only be 4, and y=1

And 41 2 = 1681That's what it's asked for!

Let the first two digits be x, and the last two digits are y, according to the meaning of the title, you can get Wang Xiao:

x+y）² 100x +y

Collation, we get: x +2y - 100) x +y -y = 0 This trapped virtual equation has a positive integer solution, so 10000 - 396 y 0 and is a perfect square number.

y 25 when y = 25, 100, then x = 30 or 20 Therefore, such four digits can be: 3025, 2025

Let the thousands, hundreds, and digits be a, and the tens, and the single digits are b

a×1000＋a×100＋b×10＋b

a×1100＋b×11

11×（a×100＋b）

A 100 b is divisible by 11, a b 11, and the four-digit number 11 (a 100 (11 a)).

11×（a×99＋11）

11×11×（9a+1)

9a 1 is a perfectly squared number.

Bring A into 9A+1

9a is true only when a 7 and b 4 are true.

The four-digit number is 7744

Four digits can be expressed as:

a×1000＋a×100＋b×10＋b

a×1100＋b×11

11×（a×100＋b）

Since a 100 b must be divisible by 11, a b 11, which is brought into the above equation to give the four-digit number 11 (a 100 (11 a)) 11 (a 99 11).

11×11×（9a+1)

As long as 9a 1 is a perfect square number, it's fine.

Verified by a, 9a.

So there is only one solution for a 7; b＝4。

Therefore, the four-digit number is 7744 11 2 8 2 = 88 88.

Let the first two digits and the last two digits of the four digits be a, b respectively, and (a+b) 2=100a+b according to the title

a=50-b±√(2500-99b)

Since a and b are integers, (2500-99b) are integers, i.e. 2500-99b are perfectly squared.

Using the exhaustive method to get b can only be 1 and 25

So a is 98, 20, 30

So this four-digit number is 2025, 3025, 9801

Let the first two digits be x and the last two digits be y, according to the title:

x+y) Honor Sparrow =100x +y

solution, so 10000 - 396 y 0 and is perfectly squared.

y Sleepy Fiction 25

When y = 25, 100, then x = 30 or 20 Therefore, such a four-digit number Wang Xiao can be: 3025, 2025

2025＝（20+25）^2

A stupid method, according to the title, this four-digit number can be squared, so the minimum of this number may be 1024 32 squared and so on (it would be better if there was a calculator or square table):

Let the first two digits be x and the last two digits be y, according to the title there are the following conditions (x+y) 2=100x+y

y belongs from 0 to 99

x belongs to 10 to 99

x+y) 2 belongs to 1000 to 9999

x,y are integers.

You can find the point that satisfies the condition and find the result 2025 3025 9801

Solution: Let the first two digits be x and the last two digits be y, according to the meaning of the question

x+y）² = 100x +y

Collation, we get: x +2y - 100) x +y -y = 0 This equation has a positive integer solution, so = 10000 - 396 y 0 and is a perfect square number.

y 25 when y = 25, 100, then x = 30 or 20 Therefore, such four digits can be: 3025, 2025

Solution: Because it is a perfect square number, you can set it to k square, then there is k*k = 1100a + 11b, so k*k must be a multiple of 11, so k must be a multiple of 11, only 33, 44, 55, 66, 77, 88, 99 are left, 1100a + 11b 11 = >100a + b

100a + b is written as a0b, it must also be a multiple of 11, a0b 11 where ten digits can only be a-1, then the remainder is 11-a, and b must be the same as this remainder, because (11-a)b is divisible by 11, so a+b = 11, then only 2299, 3388, 4477, 5566 (high and low can be exchanged).

Obviously, 33 and 44 were immediately excluded, because the minimum was also 2299, which could not be reached, so only 55, 66, 77, 88, 99 squared remained, and the square meter of 99 ended with 1, so it was immediately excluded, leaving only 55, 66, 77, 88 to make up 2299, 3388, 4477, 5566 (high and low exchangeable).

The square of 55 must be a multiple of 25, so it is excluded.

Only 66,77,88 are left to make up 2299,3388,4477,5566 (high and low exchangeable).

The square of 66 must be a multiple of 9, but a+b = 11, so it is not divisible by 9.

So it can only be 77 or 88, if it is 77, then it must be 2299, because the end of the square of 7 must be 9, obviously not right, then only 88 is left, 88 * 88 = 90-2) * (90-2) = 8100 - 4 * 90 + 4 = 8104 - 360 = 7744

So the number is 7744