Question

1.В комнате стоят несколько четырехногих и несколько трехногих табуретов. Когда на всех стульях и табуретках сидят по человеку, в комнате 39 ног. Сколько в комнате стульев и сколько табуретов.. решение... 2. Доказать что число ababab делится на 21.

Answer 1

Answer:

1. В комнате стоят несколько четырехногих и несколько трехногих табуретов. Когда на всех стульях и табуретках сидят по человеку, в комнате 39 ног. Сколько в комнате стульев и сколько табуретов?

Let's solve this problem step by step. Let's assume there are x four-legged stools and y three-legged stools in the room.

Each four-legged stool has 4 legs, and each three-legged stool has 3 legs. Since there are no other objects in the room, the total number of legs can be calculated as follows:

Total number of legs = (number of four-legged stools * 4) + (number of three-legged stools * 3)

Given that the total number of legs in the room is 39, we can set up the equation:

4x + 3y = 39

To find the number of stools, we need to find the values of x and y that satisfy this equation.

Let's solve this equation using a method called substitution:

From the equation 4x + 3y = 39, we can isolate x in terms of y:

4x = 39 - 3y x = (39 - 3y) / 4

Now, let's substitute this value of x into the equation 4x + 3y = 39:

4((39 - 3y) / 4) + 3y = 39

Simplifying the equation:

39 - 3y + 3y = 39 39 = 39

This equation is true for any value of y. Therefore, there are infinitely many solutions to this problem. The number of four-legged stools (x) and three-legged stools (y) can vary as long as the total number of legs is 39.

2. Доказать, что число ababab делится на 21.

To prove that the number ababab is divisible by 21, we need to show that it is divisible by both 3 and 7, as 21 is the product of these two prime numbers.

Let's break down the number ababab:

ababab = 100001a + 10010b = 100001a + 10010b + 1000(a - b) + 100(a - b) = 1001(100a + 10b + a - b)

Now, let's simplify the expression:

ababab = 1001(101a + 9b)

Since 1001 is divisible by both 3 and 7, we need to show that (101a + 9b) is also divisible by both 3 and 7.

To prove divisibility by 3, we can check if the sum of the digits is divisible by 3. In this case, (101a + 9b) has the same digits as ababab, so the sum of the digits is:

a + b + a + b + a + b = 3a + 3b

Since 3a + 3b is divisible by 3, (101a + 9b) is also divisible by 3.

To prove divisibility by 7, 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.

In this case, the units digit is b, and the remaining digits are 101a + 9b. So, the difference is:

2b - (101a + 9b) = -101a - 7b

Since -101a - 7b is divisible by 7, (101a + 9b) is also divisible by 7.

Therefore, since (101a + 9b) is divisible by both 3 and 7, the number ababab is divisible by 21.

Please let me know if you have any further questions!