We can change a natural number $n$ in three ways: a) If the number $n$ has at least two digits, we erase the last digit and we subtract that digit from the remaining number (for example, from $123$ we get $12 - 3 = 9$); b) If the last digit is different from $0$, we can change the order of the digits in the opposite one (for example, from $123$ we get $321$); c) We can multiply the number $n$ by a number from the set $ \{1, 2, 3,..., 2010\}$. Can we get the number $21062011$ from the number $1012011$?
Problem
Source: JBMO 2011 Shortlist C3
Tags: JBMO, combinatorics
RagvaloD
14.10.2017 11:13
Let $11|N$ then every operation give new number $M$ with $11|M$ a) $N=10A+B \equiv B-A=0 \pmod {11}$ and we get $A-B \equiv 0 \pmod 11$ So after operatioon a) we get number that is divisible 11 too b)Easy to show if $N$ is divisible $11$ then new number is divisible too. c)obvious $11|1012011$ but $11 \not |21062011$ so answer NO
capristogui
26.11.2021 00:40
consider: 21062011 $\rightarrow$ (a) 2106201-1 = 2106200 $\rightarrow$ (a) 210620-0 $\rightarrow$ (a) 21062 $\rightarrow$ 2106-2 = 2104 $\rightarrow$ (a) 210 - 4 = 206 $\rightarrow$ (a) 20-6 = 14 $\rightarrow$ (b) 41 $\rightarrow$ (a) 4-1 = 3, now, multiply that by 13, 11, 337, 7 and we get 3 $\cdot$ 13 $\cdot$ 11 $\cdot$ 337 $\cdot$ 7 = 1012011, and we are done.
sixoneeight
26.11.2021 01:43
you went backwards...
capristogui
27.11.2021 23:43
sorry. I translated wrong