Solution

(A) Yes, for example, as follows:

\[17 \rightarrow 34 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 1\]

**(B)** No, it is not possible. The initial number \(17\) is not divisible by \(3\). If a number is not divisible by \(3\), then after performing the allowed operations the resulting number will not be divisible by \(3\): - if a number not divisible by \(3\) is multiplied by \(2\), the product is also not divisible by \(3\); - if from a number not divisible by \(3\) one deletes a digit \(0,3,6,9\), then the resulting number is also not divisible by \(3\) (the digit sum of the initial number is not divisible by \(3\) (the divisibility rule for \(3\)); deleting \(0,3,6,9\) yields a digit sum that is still not divisible by \(3\)). Therefore, even after several operations, the resulting number will not be divisible by \(3\). This means that the number \(15\) cannot be obtained, since it is divisible by \(3\).

Atrisinājums

(A) Jā, var, piemēram, šādi:

\[17 \rightarrow 34 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 1\]

**(B)** Nē, nevar. Sākotnējais skaitlis \(17\) nedalās ar \(3\). Ja skaitlis nedalās ar \(3\), tad, izpildot dotās darbības, iegūtais skaitlis nedalīsies ar \(3\): - ja skaitli, kas nedalās ar \(3\), reizina ar \(2\), tad arī iegūtais reizinājums nedalīsies ar \(3\); - ja skaitlim, kas nedalās ar \(3\), izsvītro ciparu \(0,3,6,9\), tad arī iegūtais skaitlis nedalīsies ar \(3\) (sākotnējā skaitla ciparu summa nedalās ar \(3\) (dalāmības pazīme ar \(3\)), ja izsvītros \(0,3,6,9\), tad arī iegūs ciparu summu, kas nedalās ar \(3\)). Tātad arī pēc vairākām operācijām iegūtais skaitlis nedalīsies ar \(3\). Tas nozīmē, ka skaitli \(15\) iegūt nevar, jo tas dalās ar \(3\).