Question

Вася написал на доске число 13579 несколько раз подряд. Петя заявил, что полученное число делится на 2019. Может ли Петя быть прав?​

Answer 1

Число 13579 не делится на 3.

1+3+5+7+9=25. 25 не делится на 3, а значит и 13579 тоже не делится.

Число 2019 делится на 3.

2+0+1+9=12. 12 делится на 3, а значит и 2019 тоже делится.

Если 13579 не делится на 3, в то время как 2019 делится на 3, то 13579 не получится поделить на 2019. Вывод: Петя не прав.

Answer 2

To determine if the number 13579 repeated multiple times is divisible by 2019, we need to find the remainder when dividing this number by 2019.

Let's calculate:

The number 13579 is formed of the digits 1, 3, 5, 7, and 9, and its length is 5. So, when this number is repeated multiple times, it can be represented as:

13579 * 10^0 + 13579 * 10^1 + 13579 * 10^2 + ... + 13579 * 10^n

where n is the number of repetitions. Since each term has a factor of 10, we can factor out 13579 from the sum:

13579 * (10^0 + 10^1 + 10^2 + ... + 10^n)

Now, the sum of powers of 10 (1, 10, 100, 1000, ...) can be calculated as follows:

10^0 + 10^1 + 10^2 + ... + 10^n = (10^(n+1) - 1) / (10 - 1) = (10^(n+1) - 1) / 9

Now we can rewrite the number as:

13579 * ((10^(n+1) - 1) / 9)

To check if this number is divisible by 2019, we need to check if it leaves a remainder of 0 when divided by 2019:

13579 * ((10^(n+1) - 1) / 9) ≡ 0 (mod 2019)

Now, we know that 2019 = 3 * 673.

Since 3 and 673 are both prime numbers, we can split the congruence into two parts:

13579 * ((10^(n+1) - 1) / 9) ≡ 0 (mod 3)

13579 * ((10^(n+1) - 1) / 9) ≡ 0 (mod 673)

Now, 13579 is not divisible by 3, so we only need to focus on the second part.

To find if ((10^(n+1) - 1) / 9) is divisible by 673, we can check if (10^(n+1) - 1) is divisible by 9 and 673, because then the whole expression will be divisible by 673.

10^(n+1) - 1 ≡ 0 (mod 9)

10^(n+1) - 1 ≡ 0 (mod 673)

The first congruence is satisfied when (n+1) is divisible by 6, as 10^6 ≡ 1 (mod 9). So, n must be of the form n = 6k - 1, where k is an integer.

The second congruence is satisfied when (n+1) is divisible by 672, as 10^672 ≡ 1 (mod 673). So, n must be of the form n = 672k - 1, where k is an integer.

Now, let's find the common solution for n using the Chinese Remainder Theorem (CRT) as 6 and 672 are relatively prime:

n = 6k - 1 = 672m - 1

Simplifying: 6k = 672m

Now, it's apparent that the smallest positive value of k and m that satisfies this equation is k = 112 and m = 1. Therefore, the smallest value for n is:

n = 6 * 112 - 1 = 671

Now, if we repeat the number 13579, 671 times, we will have a number that is divisible by 2019. Therefore, Petya can be right. The number formed by repeating 13579, 671 times, is divisible by 2019.