The product of three consecutive integers is divisible bya)5b)6c)7d)none of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

- Multiples of three appear after a gap of two consecutive integers. For example, if the a number 'a' is divisible by 3 then (a+3), (a+6),etc are all multiple of threes. In other words there will be exactly two numbers between any consecutive multiples of three. Also, in any two consecutive integers one will be a multiple of two.

- If you take any three consecutive integers, there will be at-least one even number and one multiple of three. Any even number is of the form 2k and any multiple of three is of the form 3m. Let the other number be n. So, the product will be (2k)(3m)(n) = 6kmn => the product is always divisible by 6

Answer

6 ,option b

Answer

Suppose first integer be 1
Second = 1 + 1 = 2
Third = 1 + 2 = 3
Product = ( 1) × ( 2 ) × ( 3 ) = 6 which is divisible by 6
You can also suppose other integer in place of 1
Let consider again first integer be = 2
Second consecutive integer = 2 +1 = 3
Third = 2 + 2 = 4
Product = ( 2 ) × ( 3 ) × ( 4 )
= 24 which is divisible by 6
Hence , option (B) is correct .

Answer

**Solution:**

To solve this problem, let's assume the three consecutive integers as $n$, $n+1$, and $n+2$.

We need to determine if the product of these three consecutive integers is divisible by 6.

**Divisibility by 6:**

For a number to be divisible by 6, it must be divisible by both 2 and 3.

**Divisibility by 2:**

Every even number is divisible by 2. In our case, $n$ can be either even or odd.

- If $n$ is even, then $n+1$ is odd and $n+2$ is even. So, we have two even numbers in the product, ensuring that it is divisible by 2.
- If $n$ is odd, then $n+1$ is even and $n+2$ is odd. However, in this case, we have only one even number in the product, which means it is not divisible by 2.

Therefore, in order for the product of three consecutive integers to be divisible by 6, $n$ must be an even number.

**Divisibility by 3:**

For a number to be divisible by 3, the sum of its digits must be divisible by 3.

Let's consider the three consecutive integers $n$, $n+1$, and $n+2$:

- If $n$ is divisible by 3, then the sum of its digits is divisible by 3.
- If $n+1$ is divisible by 3, then the sum of its digits is divisible by 3.
- If $n+2$ is divisible by 3, then the sum of its digits is divisible by 3.

Therefore, at least one of the three consecutive integers will be divisible by 3, ensuring that the product is divisible by 3.

**Combining Divisibility by 2 and 3:**

We have shown that the product of three consecutive integers is divisible by 2 and 3.

Since 2 and 3 are both factors of 6, the product is also divisible by 6.

Hence, the correct answer is option **B) 6**.

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