Divisibility and Multiples | FIO

Question 30

Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

Solution

The divisibility of a product of consecutive integers depends on the presence of factors of prime numbers within that sequence.

Step 1 — Product of two consecutive integers

Let us consider two consecutive integers. One integer must be even. The other integer must be odd. An even number is always a multiple of 2. So, their product will always have a factor of 2. Let the two consecutive integers be $n$ and $n+1$ .

Case 1: $n$ is an even integer. If $n$ is even, we can write $n = 2k$ for some integer $k$ . The product is $n(n+1)$ . $n(n+1) = (2k)(n+1)$ $= 2k(n+1)$ This product is clearly a multiple of 2.

Case 2: $n$ is an odd integer. If $n$ is odd, then $n+1$ must be an even integer. We can write $n+1 = 2k$ for some integer $k$ . The product is $n(n+1)$ . $n(n+1) = n(2k)$ $= 2nk$ This product is also clearly a multiple of 2. In both cases, the product of two consecutive integers is always a multiple of 2.

$\boxed{\text{Yes, the product of two consecutive integers is always a multiple of 2.}}$

Step 2 — Product of three consecutive integers

For a number to be a multiple of 6, it must be a multiple of both 2 and 3. From Step 1, we know the product of two consecutive integers is a multiple of 2. So, the product of three consecutive integers will also be a multiple of 2. Now, let us check for divisibility by 3. In any three consecutive integers, one integer must be a multiple of 3. Let the three consecutive integers be $n, n+1, n+2$ .

Case 1: $n$ is a multiple of 3. If $n$ is a multiple of 3, then the product $n(n+1)(n+2)$ is a multiple of 3.

Case 2: $n$ is not a multiple of 3. This means $n$ can be written as $3k+1$ or $3k+2$ for some integer $k$ . If $n = 3k+1$ : Then $n+2 = (3k+1)+2$ . $n+2 = 3k+3$ $= 3(k+1)$ This means $n+2$ is a multiple of 3. So, the product $n(n+1)(n+2)$ is a multiple of 3.

If $n = 3k+2$ : Then $n+1 = (3k+2)+1$ . $n+1 = 3k+3$ $= 3(k+1)$ This means $n+1$ is a multiple of 3. So, the product $n(n+1)(n+2)$ is a multiple of 3. In all cases, one of the three integers is a multiple of 3. Therefore, the product of three consecutive integers is a multiple of 3. Since the product is a multiple of both 2 and 3, it is a multiple of $2 \times 3 = 6$ .

$\boxed{\text{Yes, the product of three consecutive integers is always a multiple of 6.}}$

Step 3 — Product of four consecutive integers

The product of $k$ consecutive integers is always a multiple of $k!$ (k factorial). The factorial of a positive integer $k$ , written as $k!$ , is the product of all positive integers less than or equal to $k$ . For 4 consecutive integers, the product is a multiple of $4!$ . Let us calculate $4!$ . $4! = 4 \times 3 \times 2 \times 1$ $= 24$ So, the product of 4 consecutive integers is always a multiple of 24. This is because among any four consecutive integers, we will always find: One number that is a multiple of 4. At least one other number that is a multiple of 2. One number that is a multiple of 3. So, the product will contain factors of $4 \times 3 \times 2 = 24$ .

Step 4 — Product of five consecutive integers

For 5 consecutive integers, the product is a multiple of $5!$ . Let us calculate $5!$ . $5! = 5 \times 4 \times 3 \times 2 \times 1$ $= 120$ So, the product of 5 consecutive integers is always a multiple of 120. This is because among any five consecutive integers, we will always find: One number that is a multiple of 5. One number that is a multiple of 4. One number that is a multiple of 3. At least one other number that is a multiple of 2. So, the product will contain factors of $5 \times 4 \times 3 \times 2 = 120$ .

Answer

(i) Yes, the product of two consecutive integers is always a multiple of 2. This is because one of the two consecutive integers must be an even number, which is a multiple of 2. (ii) Yes, the product of three consecutive integers is always a multiple of 6. This is because among any three consecutive integers, there is always at least one multiple of 2 and at least one multiple of 3. Since 2 and 3 are coprime, their product is a multiple of $2 \times 3 = 6$ . (iii) The product of 4 consecutive integers is always a multiple of 24. (iv) The product of 5 consecutive integers is always a multiple of 120.

More questions in FIO

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Q13

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Q14

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Q15

What will be the digital root of the number $9a + 36b + 13$ ?

Q16

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Q17

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Q18

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Q19

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Q20

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Q21

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Q22

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Q23

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Q24

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Q27

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Q28

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Q29

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Q30

Q31

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Q32

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