Question

4.3. Докажите, что разность между трехзначным числом и числом, записанным теми же цифрами, но в обратном порядке, не может быть квадратом натурального числа.​

Answer 1

n ∈ N

n > 0

пусть число XYZ тогда обратное ZYZ

XYZ - ZYX = 100X + 10Y + Z - 100Z - 10Y - X = 99X - 99Z = 99(X - Z)

допустим что есть такое K

что 99(X - Z) = K²  

так 99(X-Z) кратно 3 и 11 то и K кратно 3 и 11 и на3*11 = 33

значит K = 33M  M∈N

K² = (33M)² < 1000  (99*9 < 1000)

1089M² < 1000

M² < 1000/1089 < 1

противоречие M ∈ N

доказано

Answer 2

Proof that the difference between a three-digit number and the number written with the same digits in reverse order cannot be a perfect square

To prove that the difference between a three-digit number and the number written with the same digits in reverse order cannot be a perfect square, we will use a proof by contradiction.

Let's assume that there exists a three-digit number, let's call it "abc", where a, b, and c are the digits of the number. The number written with the same digits in reverse order would be "cba".

The difference between these two numbers can be expressed as:

abc - cba = 100a + 10b + c - (100c + 10b + a) = 99a - 99c = 99(a - c)

Now, let's assume that the difference 99(a - c) is a perfect square. We can express it as:

99(a - c) = k^2, where k is a natural number.

Dividing both sides of the equation by 9, we get:

11(a - c) = (k/3)^2

Since 11 is a prime number, it means that (a - c) must be divisible by 11 for the equation to hold. However, the difference between two digits cannot be divisible by 11 unless the two digits are the same.

Therefore, we can conclude that the difference between a three-digit number and the number written with the same digits in reverse order cannot be a perfect square.

Note: The proof above is based on logical reasoning and does not require external sources.