Curious about nines

I recently noticed a curious fact about the number nine and the result of multiplying it. Could you provide an explanation for this beyond "coincidence"? The curious fact is this:

If you multiply any number from 1 through 31 by nine the result is a number that when the digits are added either equals nine or results in two nines. For example, 9 x 2 = 18; and 9 x 31 = 279.

When you continue multiplying 9 by larger and larger numbers a pattern appears to emerge.

In the 30's, only one number, 32, does not result in a sum that supports this phenomenon. 32 x 9 = 288. But 33 x 9 = 297.

In the 40's, only two numbers do not work, 42 and 43.

In the 50's, three numbers don't work, 52, 53, 54.

In the 60's, four numbers don't work, 62-65.

As you might expect, in the 70's five numbers, 72-76, don't work.

I expect this pattern goes on and then perhaps doubles back. While it has been an amusing dinner time conversation, I have not checked it out.

Is this explicable?

Jon Metropoulos

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It's simpler to analyse this problem if you repeat the digit addition whenever it fails, and replace all 9's by zeroes whenever they appear. All multiples of 9 will then yield the answer 0.

So, $$\text{Unknown environment 'tabular'}$$ \par This calculation actually succeeds in calculating the remainder when a number is divided by 9. So for example... $$\text{Unknown environment 'tabular'}$$ and 7643/9 = 849 remainder 2. \par Using this modified procedure (known to accountants as 'casting out nines'), you can get a quick check on the accuracy of any sum by doing the equivalent sum with all nines cast out (i.e. in arithmetic modulo 9). \par So... does 26 + 389 + 35 + 36 = 498? \par No! because $$(26+389+35+36)mod9=(8+2+8+0)mod9=0$$ but $$498mod9=3$$ \par I'll leave proving why repeatedly taking the sum of the digits is equivalent to finding the remainder modulo 9 to you. It all depends on noticing that any number N expressed in base 10 can be written as: $$\text{Unknown environment 'displaymath'}$$ where the $d_j$ are the digits of the number, and noting that 10 mod 9 = 1! \par (Note that 9 is not the only digit $d$ where 10 mod $d$ = 1! Is there another divisibility test lurking here somewhere?)

Going back to your original definition of whether a number works or not, it would make sense for a pattern to emerge for small numbers, but for it to get increasingly chaotic as the numbers get larger. Is this what you find?

Mike Pearson

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Paradoxical proof

James R. Foster

Please can someone disprove this?

Since $$\text{Unknown environment 'displaymath'}$$ where $i$ = $\sqrt{-1}$ and $t$ is any angle in radians, therefore $$\begin{array}{r}\\ e^{i\pi }& =& -1+0i\\ e^{i\pi }+1& =& 0\\ i\pi & =& \ln (-1)\\ from\ln (ab)& =& \ln (a)+\ln (b)\\ \ln (-a)& =& i\pi +\ln (a)\end{array}$$ Now, from equation 1 we have: $$\begin{array}{r}\\ 0& =& \ln (-e^{i\pi })\\ 0& =& i\pi .\ln (-e)\end{array}$$ From equation 2 we have: $$\begin{array}{r}\\ 0& =& i\pi .(i\pi +1)\\ 0& =& (i^2)({\pi }^2)+i\pi \\ -i\pi & =& (-1)({\pi }^2)\\ -i& =& -\pi \\ \pi & =& i\end{array}$$ or, keeping everything real... ${\pi }^4=1$.

This can be disproven by saying "oi, look, no it doesn't", but can anyone disprove this paradoxical proof?

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The main problem is with the lines:- $$\begin{array}{r}\\ 0& =& \ln (-e^{i\pi })\\ 0& =& i\pi .\ln (-e)\end{array}$$ which does not follow as $-e^{i\pi }$ means $(-1)\times e^{i\pi }$ rather than $(-e{)}^{i\pi }$. The second problem is that the logarithm of a complex number is not a well-defined function unless you specify its range more carefully. This is because $z=r{e}^{i\theta }=r{e}^{i(\theta +2k\pi )}$ for any integer $k$. So: $$\ln (z)=\ln (r)+i(\theta +2k\pi )$$ The correct deduction from (1) is $$0=i(\pi +2k\pi )+i(\pi +2m\pi )$$ for some integral values of $k$ and $m$. ...which is correct ($k=0,m=-1$ for example)

Mike Pearson

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Mysterious sequence

Please could you help us solve a maths sequence we have been puzzled with for a while.

It's to find the Nth term of the following sequence:

0, 11, 28, 51, 80

Miss G. Cornish

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This one yields to the method of differences...

   1   2   3   4   5  = n     
                              
   0  11  28  51  80  = S(n)  
    \ / \ / \ / \ /            
     11  17  23  29              
       \ / \ / \ /                
        6   6   6                 

Since we have a constant difference in the second row, a quadratic formula for the nth term will work - something like: $$\text{Unknown environment 'displaymath'}$$

Insert some values for n and S(n) and solve the resulting equations for a, b and c and the answer should drop out.

Alternatively, you could look for patterns in the factors.

Mike Pearson

Waves

I am in the process of starting a piece of A-level coursework on mathematical equations of wave motion. I am hoping to focus on simple harmonic motion and I was hoping that you could provide me with a starting point on how to set up differential equations around s.h.m, enabling me to investigate waves to the fullest.

Amrit Summan

You will get a clear answer to this question if you simply refer to your A level mechanics text book.

Gyroscopes

Can you provide a source for the battery-powered gyros you mention [in the Issue 7 article Galloping gyroscopes]?

Dwight Ufford

These were custom made in the University Engineering Lab and we think they would be quite expensive to manufacture. You could try the gyro car enthusiasts - they may be able to point you to a lower power version suitable for hand-held use.

Constructive criticism?

"this is one of the worst websites ive ever been on ever and its boring and geeky and dull and stupid and im only on cos i was forced on it would never encourage someone in to mathss" [sic]

Anonymous Berkshire student

How to contact us: Any comments?