Question

Натуральное число делится на 42. Сумма цифр, не участвующих в написании этого числа, равна 35. Докажите, что в записи числа есть хотя бы две одинаковые цифры

Answer 1

Поскольку сумма всех цифр от 0 до 9 равна:

0+1+2+...+9 = 9*10/2 = 45, то сумма всех элементов множества цифр, входящих в состав данного числа равна: 45 - 35 = 10.

Предположим, что в данном числе все цифры различны, но тогда все его цифры совпадают со всеми элементами данного множества, иначе говоря, сумма цифр этого числа равна 10, но тогда, поскольку сумма цифр данного числа не делится на 3 ( 10 не делится на 3), то и данное число не делится на 3, но по условию данное число делится на 42, а значит должно делится на 3, то есть мы пришли к противоречию.

Как видим, хотя бы две цифры данного числа должны быть одинаковыми.

Answer 2

Problem Analysis

We are given a natural number that is divisible by 42. The sum of the digits that are not used in writing this number is 35. We need to prove that there are at least two identical digits in the representation of this number.

Proof

Let's assume that there are no two identical digits in the representation of the given number. This means that each digit from 0 to 9 appears at most once in the number.

Since the number is divisible by 42, it must also be divisible by 2, 3, and 7, which are the prime factors of 42.

Divisibility by 2: If a number is divisible by 2, its last digit must be even. Since each digit from 0 to 9 appears at most once in the number, there must be at least one even digit in the number.

Divisibility by 3: If a number is divisible by 3, the sum of its digits must be divisible by 3. The sum of the digits that are not used in writing the number is given as 35. Since each digit from 0 to 9 appears at most once in the number, the sum of the digits in the number must be 45 (0+1+2+3+4+5+6+7+8+9). Therefore, the sum of the digits in the number is divisible by 3.

Divisibility by 7: There is no direct rule for divisibility by 7 based on the sum of digits. However, we can use the divisibility rule for 7, which states that if the difference between twice the units digit and the remaining digits is divisible by 7, then the number is divisible by 7.

Let's assume that the units digit of the number is x. Since each digit from 0 to 9 appears at most once in the number, the remaining digits can be represented as 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 - x. The difference between twice the units digit and the remaining digits is:

2x - (0+1+2+3+4+5+6+7+8+9-x) = 2x - 45 + x = 3x - 45

For the number to be divisible by 7, 3x - 45 must be divisible by 7. This means that 3x must be congruent to 45 modulo 7.

Using the given search results, we can find that 3 is congruent to 3 modulo 7 Therefore, 3x must be congruent to 3 * x modulo 7.

Let's consider all possible values of x from 0 to 9:

- For x = 0, 3 * x = 0, which is not congruent to 3 modulo 7. - For x = 1, 3 * x = 3, which is congruent to 3 modulo 7. - For x = 2, 3 * x = 6, which is congruent to 6 modulo 7. - For x = 3, 3 * x = 9, which is congruent to 2 modulo 7. - For x = 4, 3 * x = 12, which is congruent to 5 modulo 7. - For x = 5, 3 * x = 15, which is congruent to 1 modulo 7. - For x = 6, 3 * x = 18, which is congruent to 4 modulo 7. - For x = 7, 3 * x = 21, which is congruent to 0 modulo 7. - For x = 8, 3 * x = 24, which is congruent to 3 modulo 7. - For x = 9, 3 * x = 27, which is congruent to 6 modulo 7.

From the above analysis, we can see that only for x = 1 and x = 7, 3x is congruent to 3 modulo 7. Therefore, the units digit of the number must be either 1 or 7.

Now, let's consider the sum of the digits that are not used in writing the number. We are given that this sum is 35. Since each digit from 0 to 9 appears at most once in the number, the sum of the digits in the number must be 45. Therefore, the sum of the digits that are used in writing the number is 45 - 35 = 10.

Since the units digit of the number must be either 1 or 7, and the sum of the digits that are used in writing the number is 10, the remaining digits must sum up to either 9 or 3.

If the remaining digits sum up to 9, then the possible digits are 2 and 6. However, we have already shown that the units digit must be either 1 or 7. Therefore, the remaining digits must sum up to 3.

The only possible digits that sum up to 3 are 0 and 3. However, we have already shown that the units digit must be either 1 or 7. Therefore, it is not possible for the remaining digits to sum up to 3.

Since we have reached a contradiction, our initial assumption that there are no two identical digits in the representation of the given number must be false. Therefore, we can conclude that there are at least two identical digits in the representation of the given number.

Conclusion

We have proven that if a natural number is divisible by 42 and the sum of the digits that are not used in writing the number is 35, then there must be at least two identical digits in the representation of the number.