Proof a Number is Divisible by 9

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In summary, to show that n^3(n+1)^3(n+2)^3 is divisible by 9, you can use the fact that every third integer is a multiple of 3, and therefore one of the three consecutive integers n, (n+1), (n+2) will be a multiple of 3. This means that its cube will be divisible by 27, and therefore the product of the cubes of the three consecutive integers will also be divisible by 9. No induction is necessary.
  • #1
hammonjj
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Show that n^3(n+1)^3(n+2)^3 is divisible by 9.

The claim is obvious, so I tried to solve by induction, but I don't see how to simplify it so the two side are equal. You end up with:

(n+1)^3(n+2)^3(n+3)^3

Thanks in advance!
 
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  • #2
hammonjj said:
Show that n^3(n+1)^3(n+2)^3 is divisible by 9.

The claim is obvious, so I tried to solve by induction, but I don't see how to simplify it so the two side are equal. You end up with:

(n+1)^3(n+2)^3(n+3)^3

Thanks in advance!

If f(k+1) - f(k) is a multiple of 9, you've done the inductive step.

So find the difference between those. To help you, use this nifty identity: [itex]a^3 - b^3 = (a-b)(a^2 + ab + b^2)[/itex]. Don't expand everything into a big mess. The first term (the a-b part) can be manipulated quite easily into a factor of 3. The second term (the a^2 + ab + b^2 part) can be factorised into 3*something after a bit of manipulation. So the whole expression becomes 9*something, and you're done (apart from showing it for n=1, which is trivial).
 
  • #3
In fact, you don't even need induction. What can you say about three consecutive integers n, (n+1), (n+2)? Would one of them necessarily be a multiple of 3? Every third integer is a multiple of 3, right?

You can immediately see that the cube of one of those is divisible by 27 (and hence by 9).
 
  • #4
Curious3141 said:
In fact, you don't even need induction. What can you say about three consecutive integers n, (n+1), (n+2)? Would one of them necessarily be a multiple of 3? Every third integer is a multiple of 3, right?

You can immediately see that the cube of one of those is divisible by 27 (and hence by 9).

I think my reading skills are failing me. Can you explain this a different way? I understand that every third n is divisible by three, just not the second part or how I could write that in proof form.

Thanks!
James
 
  • #5
hammonjj said:
I think my reading skills are failing me. Can you explain this a different way? I understand that every third n is divisible by three, just not the second part or how I could write that in proof form.

Thanks!
James

The fact that out of three consecutive integers, one is a multiple of 3 needs no proof (it's just a consequence of the three times multiplication table!).

So that integer can be expressed as 3k. Its cube is 27k^3. It is therefore divisible by 9 (since 27 = 9*3), and the product of the cubes of those three integers is also obviously divisible by 9.

You can, of course, make an even stronger proof that the product of the cubes is divisible by 27, but the question only asked for divisibility by 9. :smile:
 

What is the rule for determining if a number is divisible by 9?

The rule for determining if a number is divisible by 9 is to add up all the digits in the number, and if the sum is divisible by 9, then the number is also divisible by 9.

What is the divisibility test for 9?

The divisibility test for 9 is to add up all the digits in the number, and if the sum is divisible by 9, then the number is also divisible by 9.

How can I use the divisibility rule for 9 to check if a larger number is divisible by 9?

To check if a larger number is divisible by 9 using the divisibility rule, you can keep adding up the digits until you are left with a single digit. If that digit is divisible by 9, then the original number is also divisible by 9.

What happens if the sum of the digits is not divisible by 9?

If the sum of the digits in a number is not divisible by 9, then the number is not divisible by 9.

Can a number be divisible by 9 if the sum of its digits is 9?

Yes, a number can be divisible by 9 if the sum of its digits is 9. For example, 18 is divisible by 9 even though the sum of its digits is 9.

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