Factors of Consecutive Integers | Arithmetic

Let’s start today’s post with some magic tricks:

Trick 1

Step 1: Pick any two consecutive integers (Don’t tell me what they are!).

Step 2: Multiply them.

Whatever your numbers, the product is an even number! If you are wondering how I knew that, you desperately need to read this post. If you are shaking your head in disappointment, wait, I have more.

Note: 0 is even. So are -2, -4, -6 etc

Trick 2

Step 1: Pick any three consecutive integers.

Step 2: Multiply them.

Whatever your numbers, the product is a multiple of 6. Now, if you are surprised, great, go ahead and read this post. If you are still bewildered that why the heck am I calling ‘simple Math’ ‘magic tricks,’ wait, give me one last chance.

Trick 3

Step 1: Pick any four consecutive integers.

Step 2: Multiply them.

Whatever your numbers, the product is definitely a multiple of 24. I hope I have caught your interest now. If you are still not surprised, tell me the greatest number which is definitely a factor of the product of any five consecutive integers. If you do get an answer, scroll down to the bottom of the post to check it. If you get the correct answer, you don’t need to read this post. Hopefully, I will see you next week with something more challenging. If not, then stick around.

So how did I know the numbers which were definitely factors of consecutive integers? Let me explain you the simple Math involved.

Theory:

Pick any two consecutive numbers. e.g. (5, 6) or (101, 102) or (999, 1000) etc ..

Do you agree that one of them, and only one of them will be even? Every alternate number has 2 as a factor so no matter which two consecutive numbers you pick, one of them will definitely have 2 as a factor and the other will not.

Notice a few things about integers:
…-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16…

– Every number is a multiple of 1

– Every second number is a multiple of 2

– Every third number is a multiple of 3

– Every fourth number is a multiple of 4 and so on…

If we pick any 2 consecutive integers, one and only one of them will be a multiple of 2: e.g. pick 4, 5 (4 is a multiple of 2) or pick 11, 12 (12 is a multiple of 2) etc.

If we pick any 3 consecutive integers, at least one of them will be a multiple of 2 e.g. pick 3, 4 and 5 (4 is a multiple of 2) or pick 6, 7 and 8 (6 and 8 are multiples of 2) etc. Of the 3 numbers, exactly one will be a multiple of 3 e.g. pick 3, 4 and 5 (3 is a multiple of 3) or pick 6, 7 and 8 (6 is a multiple of 3) etc. Hence the product of the 3 consecutive integers will be a multiple of 2 and 3 and therefore, of 6.

If we pick any 4 consecutive integers, two of them will be multiples of 2 e.g. pick 3, 4, 5 and 6 (4 and 6 are multiples of 2) or pick 6, 7, 8 and 9 (6 and 8 are multiples of 2) etc. Of the 4 numbers, at least one will be a multiple of 3 e.g. pick 3, 4, 5 and 6 (3 and 6 are multiples of 3) or pick 5, 6, 7 and 8 (6 is a multiple of 3). Also, exactly one number will be a multiple of 4 e.g. pick 3, 4, 5 and 6 (4 is a multiple of 4) or pick 5, 6, 7 and 8 (8 is a multiple of 4) etc. Hence the product will be a multiple of at least 4, 3 and 2 (from the even integer which is not a multiple of 4) and therefore, of 24.

Hope the concept is clear to you. Let’s look at a question now.

Question: Given that $n$ is any integer such that ${ (n-1)\timesn\times(n+1)}$ is divisible by 24, which is the following must be true?

1. Either ${n}$ is divisible by 8 or ${ (n+1)}$ is divisible by 4

2. Either ${n}$ or ${ (n^2 - 1)}$ is divisible by 3

3. ${n}$ is not divisible by 16

Solution:

$(n-1)$ , $n$ and $(n+1)$ are consecutive integers. Either $(n-1)$ and $(n+1)$ are even or $n$ is even. We know that $n$ has 8 and 3 as factors. If $n$ is even, it will be a multiple of 8 since $(n-1)$ and $(n+1)$ will be odd. If $(n-1)$ and $(n+1)$ are even, at least one of them will be a multiple of 4 (to get 8 as the factor of their product). Either $(n-1)$ or $(n+1)$ could be the integer divisible by 4. Hence statement 1 is not necessarily true.

Since the product is divisible by 24, one of $(n-1)$ , $n$ and $(n+1)$ must be divisible by 3. Either $n$ or $(n-1)(n+1) = (n^2 - 1)$ must be divisible by 3. Hence statement 2 must be true.

If the three consecutive integers are 15, 16 and 17, $n$ could be divisible by 16. All we know is that the product is divisible by 24, it could be divisible by 48 or higher numbers. So statement 3 is not necessarily true.

Answer to the 5 consecutive integers question: 120 (Shortcut – If you need 5 consecutive integers, take 1, 2, 3, 4 and 5 – their product is 120. This will be the maximum number which will definitely be a factor of any five consecutive integers.)

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