Exercițiul 104

E.104. Vom spune că un număr natural de patru cifre este echilibratechilibrat dacă prima sau ultima sa cifră este egală cu suma celorlalte cifre ale sale. Dacă abcd‾\overline{abcd} și abcd‾+1\overline{abcd}+1 sunt numere echilibrate, atunci suma a+da+d este egală cu:

a) 1515

b) 1010

c) 1111

d) 1313

e) 1414

Olimpiadă, etapa județeană, 2021
Răspuns | Soluție

Răspuns: e) 1414

Soluție:

Caz 1\bold{Caz \space 1}: a=da=d
abcd‾−ech.⇒b=c=0\hspace*{2em} \overline{abcd} - ech. \textcolor{red}{\Rightarrow} b=c=0
abcd‾+1−ech.⇒\hspace*{2em} \overline{abcd} + 1 - ech. \textcolor{red}{\Rightarrow}
∙ d<9⇒a00d‾+1=a00(d+1)‾\hspace*{2em}\bullet \space d<9 \textcolor{red}{\Rightarrow} \overline{a00d} + 1 = \overline{a00(d+1)} - neech.
∙ d=9⇒a00d‾+1=9010)‾\hspace*{2em}\bullet \space d=9 \textcolor{red}{\Rightarrow} \overline{a00d} + 1 = \overline{9010)} - neech.

Caz 2\bold{Caz \space 2}: a>d⇒d<9, b+c≥1a \gt d \textcolor{red}{\Rightarrow} d \lt 9, \space b+c \geq 1
abcd‾−ech.⇒a=b+c+d\hspace*{2em} \overline{abcd} - ech. \textcolor{red}{\Rightarrow} a=b+c+d
abcd‾+1−ech.⇒abc(d+1)‾−ech.⇒\hspace*{2em} \overline{abcd} + 1 - ech. \textcolor{red}{\Rightarrow} \overline{abc(d+1)} - ech. \textcolor{red}{\Rightarrow}
∙ a=b+c+d+1⇒a=a+1\hspace*{2em}\bullet \space a=b+c+d+1 \textcolor{red}{\Rightarrow} a=a+1 - nu conv.
∙ d+1=a+b+c⇒d≥a\hspace*{2em}\bullet \space d+1 = a+b+c \textcolor{red}{\Rightarrow} d \geq a - contradicție

Caz 3\bold{Caz \space 3}: a<d⇒a<9, b+c≥1a \lt d \textcolor{red}{\Rightarrow} a \lt 9, \space b+c \geq 1
abcd‾−ech.⇒d=a+b+c (1)\hspace*{2em} \overline{abcd} - ech. \textcolor{red}{\Rightarrow} d=a+b+c \nobreakspace \textcolor{red}{(1)}
abcd‾+1−ech.⇒\hspace*{2em} \overline{abcd} + 1 - ech. \textcolor{red}{\Rightarrow}
∙ d<9⇒abcd‾+1=abc(d+1‾\hspace*{2em}\bullet \space d<9 \textcolor{red}{\Rightarrow} \overline{abcd}+1 = \overline{abc(d+1}
∙ a=b+c+d+1⇒a>d\hspace*{4em}\bullet \space a=b+c+d+1 \textcolor{red}{\Rightarrow} a \gt d - contrad.
∙ d+1=a+b+c⇒d+1=d→\hspace*{4em}\bullet \space d+1 = a+b+c \textcolor{red}{\Rightarrow} d+1=d \rightarrow n.c.
∙d=9⇒abcd‾+1=ab(c+1)0‾⇒\hspace*{2em}\bullet \boxed{d=9} \textcolor{red}{\Rightarrow} \overline{abcd}+1 = \overline{ab(c+1)0} \textcolor{red}{\Rightarrow}
a=b+c+1 (2)\hspace*{4em} a=b+c+1 \nobreakspace \textcolor{red}{(2)}
\hspace*{4em}Din (1) și (2) ⇒d=a+a−1⇒\textcolor{red}{\Rightarrow} d=a+a-1 \textcolor{red}{\Rightarrow}
2a=9+1⇒a=5\hspace*{4em} 2a=9+1 \textcolor{red}{\Rightarrow} \boxed{a=5}

Deci d+a=9+5=14.d+a=9+5=14.

Observație: din (1) ⇒b+c=d−a=9−5=4\textcolor{red}{\Rightarrow} b+c=d-a=9-5=4, deci numărul nostru va fi de forma 5bc9‾,\overline{5bc9}, unde b+c=4,b+c=4, adică:
{5409,5319,5229,5139,5049}.\{5409, 5319, 5229, 5139, 5049\}.