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Take any $2$ digit number having different digits. Now add the bigger digit in that number. By continuing this process, you will get a multiple of $11$ i.e. both of the digits will be equal of a resulting number. For example, taking number $57$: $$57+7=64\to64+6=70\to70+7=77=11\cdot7$$ Note: if the number after addition becomes three digit number, then take the last two digits. $$89+9=98\to98+9=107\to07+7=14\to07+7=14\to14+4=18\to18+8=26\to26+6=32\to32+3=35\to35+5=40\to40+4=44=11\cdot4$$ How to prove this?

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  • $\begingroup$ How many times are you allowed to do this? It seems like you do this until you get a multiple of 11 - so the real question is whether the process ever terminates, right? $\endgroup$ – Prahlad Vaidyanathan Apr 20 '14 at 9:27
  • $\begingroup$ I don't see how you get $44$ out of $07$. $\endgroup$ – Jack M Apr 20 '14 at 9:30
  • $\begingroup$ Yes. But the process will not end we can continue it with multiple of 11 even after we will get some another multiple of 11 and process will continue. We have to prove that definitely there will a multiple of 11 for any two digit number. $\endgroup$ – Satvik Mashkaria Apr 20 '14 at 9:32
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    $\begingroup$ 07+7=14, 14+4=18, 18+8=26, 26+6=32, 32+3=35, 35+5=40, 40+4=44 !! $\endgroup$ – Satvik Mashkaria Apr 20 '14 at 9:34
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    $\begingroup$ Here's a visualization of the iteration of this map over $\mathbb Z_{100}$. It seems there's two cases, some of the numbers just go straight down to $0$ in a big river (and then stay at $0$, of course), while others fall into a whirlpool and cycle forever. Sorry the graph is a bit messy, I couldn't find good layout software. $\endgroup$ – Jack M Apr 20 '14 at 11:23
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Call $f(x):\Bbb{N}\to\Bbb{N}$ the function which, starting from a number, gives out the starting number $+$ the bigger digit in that number. Now consider only $2$-digit numbers, and if you come up with a $3$-digit number, then take the last two digits...after some explanation we're ready to start the proof. Note that $f(x) \gt x$, as $f(x)=x+a,a\gt0$, so $f(x)=x+a\gt x$ This is valid if $x\neq89,91,92,93,94,95,96,97,98,99$. Thus it's relatively simple noting that starting from a number, you get higher values, until you reach one of $89,91,92,93,94,95,96,97,98,99$, which results in

  • $89\to89+9=98\to98+9=107\to07+7=14\to07+7=14\to14+4=18\to18+8=26\to26+6=32\to32+3=35\to35+5=40\to40+4=44=11\cdot4$

  • $91\to91+9=100\to00=11\cdot0$

  • $92\to92+9=101\to01+1=2\to2+2=4\to4+4=8\to8+8=16\to16+6=22=11\cdot2$

  • $93\to93+9=102\to02+2=4\to4+4=8\to8+8=16\to16+6=22=11\cdot2$

  • $94\to94+9=103\to03+3=6\to6+6=12\to12+2=14\to14+4=18\to18+8=26\to26+6=32\to32+3=35\to35+5=40\to40+4=44=11\cdot4$

  • $95\to95+9=104\to04+4=8\to8+8=16\to16+6=22=11\cdot2$

  • $96\to96+9=105\to05+5=10\to10+1=11=11\cdot1$

  • $97\to97+9=106\to06+6=12\to12+2=14\to14+4=18\to18+8=26\to26+6=32\to32+3=35\to35+5=40\to40+4=44=11\cdot4$

  • $98\to98+9=107\to07+7=14\to07+7=14\to14+4=18\to18+8=26\to26+6=32\to32+3=35\to35+5=40\to40+4=44=11\cdot4$

  • $99=11\cdot9$

So, starting from any $2$-digit number, in the sequence $x,f(x),f(f(x))...$, will appear at least one multiple of $11$

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  • $\begingroup$ very beautifull and simple.+1 $\endgroup$ – Konstantinos Gaitanas Apr 20 '14 at 10:31
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    $\begingroup$ I know it's not too elegant, but I'm in a rush for the Easter lunch... Happy Easter to all of you! $\endgroup$ – sirfoga Apr 20 '14 at 10:32
  • $\begingroup$ That is why i like mathematicians.Happy easter! $\endgroup$ – Konstantinos Gaitanas Apr 20 '14 at 10:33
  • $\begingroup$ I don't think you need a case for 89. I just moves to 98 anyway. $\endgroup$ – Henning Makholm Apr 20 '14 at 19:14
  • $\begingroup$ Yes actually I didn't mind it... $\endgroup$ – sirfoga Apr 20 '14 at 20:43

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