Question

Петя набрал в лесу грибов. Если убрать из корзинки 1 гриб, то число грибов в корзинке будет делиться на 2, если убрать 2 гриба, то их число будет делиться на 3, если убрать 3 гриба, то их число будет делиться на 4, если убрать 4 гриба, то их число будет делиться на 5, если убрать 5 грибов, то их число будет делиться на 6. Найти число грибов в корзинке, если известно, что оно двузначно..

Answer 1

Тут число по признакам делимости искать.
Набрал грибов обозначим двузначное
а•10+в =10а+в
Десятки это а, потому на 10 умножаем, единицы это в.

1)) Если (10а+в)-1 кратно 2, это четное, значит (10а+в) не четное число, так как на 2 делятся числа Четные, где вконце (0,2,4,6,8)

2)) Если (10а+в)-2, то число кратно 3, значит сумма цифр числа (10а+в)-2 делится на 3 и это нечетное число, потому что выше если (-1) четное

3)) Если (10а+в)-3, то делится на 4, на 4 делятся числа если две последние цифры 00 или делятся на 4. Тогда с (-3) будет четное число.

4)) Если (10а+в) -4, то число кратно 5. На 5 делятся числа вконце 0 или 5. С (-4) вконце 0 или 5, значит (в =4 или 9). Вначале нашли, что (10а+в) нечетное, значит (в=9).

5)) Если (10а+в) -5, то число кратно 6. Значит оно тогда должно делится на 2 и 3 одновременно, тогда число (10а+в)-5 должно быть четное и сумма цифр делится на 3.
(10а+в)-5= (10а+9)-5= 10а+4 кратно 3 и 2, ищем число а.

Снова 2))
Чтобы делилось на 3, вычитаем 2, сумма цифр 10а+в= 10а+9-2= 10а+7 тогда а может быть 2, 2+7=9; 5, 5+7=12;
8, 8+7=15; больше нет, дальше получим трехзначное число

Снова 3))
Если а=2; 5; 8. Делимость на 4.

10а+в-3= 10а+9-3=10а+6.
а=2; 10•2+6=26; не подходит;
а=5; 5•10+6=56 делится на 4;
а=8. 10•8+6=86 не делится.

Значит а=5. в=9. Число 59 грибов.

Проверяем по условию все
10а+в=10•5+9=59.

1))59-1=58. Четное 58:2=29 делится
на 2.
2)) 59-2=57. 5+7=12. 12:3=4. Делится сумма цифр на 3.
3)) 59-3=56. 56:4=14. Делится на 4.
4)) 59-4=55. 55:5=11. Делится на 5.
5)) 59-5=54. 54 четное, 5+4=9. Сумма цифр делится на 3. Значит число на 6 делится. 54:6=9.

Ответ: в корзинке 59 грибов.

Answer 2

Problem Analysis

We are given the following information: - Petya collected some mushrooms in the forest. - If we remove 1 mushroom from the basket, the number of mushrooms in the basket will be divisible by 2. - If we remove 2 mushrooms from the basket, the number of mushrooms will be divisible by 3. - If we remove 3 mushrooms from the basket, the number of mushrooms will be divisible by 4. - If we remove 4 mushrooms from the basket, the number of mushrooms will be divisible by 5. - If we remove 5 mushrooms from the basket, the number of mushrooms will be divisible by 6.

We need to find the number of mushrooms in the basket, knowing that it is a two-digit number.

Solution

Let's analyze the given information step by step:

1. If we remove 1 mushroom from the basket, the number of mushrooms in the basket will be divisible by 2. - This means the number of mushrooms is odd, as removing 1 mushroom from an even number would result in an odd number. - Therefore, the last digit of the number of mushrooms is 1, 3, 5, 7, or 9.

2. If we remove 2 mushrooms from the basket, the number of mushrooms will be divisible by 3. - The sum of the digits of a number divisible by 3 is also divisible by 3. - Since the last digit is odd, the sum of the digits excluding the last digit should be divisible by 3. - Let's consider the possible values for the last digit and calculate the sum of the remaining digits to check for divisibility by 3:

- If the last digit is 1, the remaining digits should add up to a number divisible by 3. - For example, if the number is 31, the remaining digit is 3, and it is divisible by 3. - If the number is 51, the remaining digit is 5, and it is not divisible by 3. - If the number is 71, the remaining digit is 7, and it is not divisible by 3. - If the number is 91, the remaining digit is 9, and it is divisible by 3. - If the last digit is 3, the remaining digits should add up to a number divisible by 3. - For example, if the number is 33, the remaining digit is 3, and it is divisible by 3. - If the number is 53, the remaining digit is 5, and it is not divisible by 3. - If the number is 73, the remaining digit is 7, and it is not divisible by 3. - If the number is 93, the remaining digit is 9, and it is divisible by 3. - If the last digit is 5, the remaining digits should add up to a number divisible by 3. - For example, if the number is 35, the remaining digit is 3, and it is divisible by 3. - If the number is 55, the remaining digit is 5, and it is not divisible by 3. - If the number is 75, the remaining digit is 7, and it is not divisible by 3. - If the number is 95, the remaining digit is 9, and it is divisible by 3. - If the last digit is 7, the remaining digits should add up to a number divisible by 3. - For example, if the number is 37, the remaining digit is 3, and it is divisible by 3. - If the number is 57, the remaining digit is 5, and it is not divisible by 3. - If the number is 77, the remaining digit is 7, and it is not divisible by 3. - If the number is 97, the remaining digit is 9, and it is divisible by 3. - If the last digit is 9, the remaining digits should add up to a number divisible by 3. - For example, if the number is 39, the remaining digit is 3, and it is divisible by 3. - If the number is 59, the remaining digit is 5, and it is not divisible by 3. - If the number is 79, the remaining digit is 7, and it is not divisible by 3. - If the number is 99, the remaining digit is 9, and it is divisible by 3.

- From the above analysis, we can see that the only possible values for the last digit are 1 and 9, as they result in a sum of the remaining digits that is divisible by 3.

3. If we remove 3 mushrooms from the basket, the number of mushrooms will be divisible by 4. - A number is divisible by 4 if the last two digits form a number divisible by 4. - Since the last digit can only be 1 or 9, let's consider the possible values for the last two digits and check if they are divisible by 4:

- If the last two digits are 11, they are not divisible by 4. - If the last two digits are 19, they are not divisible by 4. - If the last two digits are 91, they are not divisible by 4. - If the last two digits are 99, they are divisible by 4.

- From the above analysis, we can see that the only possible value for the last two digits is 99.

4. If we remove 4 mushrooms from the basket, the number of mushrooms will be divisible by 5. - A number is divisible by 5 if the last digit is 0 or 5. - Since the last digit can only be 1 or 9, the number of mushrooms cannot be divisible by 5.

5. If we remove 5 mushrooms from the basket, the number of mushrooms will be divisible by 6. - A number is divisible by 6 if it is divisible by both 2 and 3. - Since the number is odd, it cannot be divisible by 2. - Therefore, the number of mushrooms cannot be divisible by 6.

From the above analysis, we can conclude that the number of mushrooms in the basket is 99.

Answer

The number of mushrooms in the basket is 99.