anonymous
  • anonymous
Pick a number 1-8. multiply the number by 3. then add three. then multiply that number by three again. you will get a two digit number, those digits add to get 9. Can someone explain how?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
The key to this problem is that any two digit number that is divisible by nine has digits that add up to get to nine: 18, 27, 36, 45, 54, 63, 72, 81, 90, 99. The next step is realizing the number you're creating is divisible by nine. You know immediately that it has to be divisible by three (because you just multiplied by three to get the final number!). But is the number before that divisible by three? A bit of checking will show that it is. Here is the steps you went through to get to the final number, where \(x\) is the original number 1-8, and then the steps showing that it is divisible by nine: \[ \begin{align*} x_2 &= 3x\\ x_3 &= x_2 + 3\\ x_\textrm{final} &= 3x_3\\ &= 3(x_2 +3)\\ &= 3(3x+3)\\ &= (3)(3)(x+1)\\ x_\textrm{final}&= 9(x+1)\\ \frac{x_\textrm{final}}{9} &= x+1 \end{align*} \] Since \(x\) was an integer, \(x+1\) is as well, so you have a final number divisible by nine!
anonymous
  • anonymous
>.<.... give me a minute to take in what was just written thanks for the response by the way!
anonymous
  • anonymous
Haha, no problem. Let me know if you have questions.

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