Getting multiple of 11

Question

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?

Answer 1

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 $11$