Exercițiul 109

E.109. Suma cifrelor numărului abcd‾\overline{abcd} pentru care 4⋅abcd‾=dcba‾4 \cdot \overline{abcd}=\overline{dcba} este egală cu:

a) 1818

b) 2222

c) 1616

d) 2424

e) 2020

Olimpiadă, etapa județeană, 2021
Răspuns | Soluție

Răspuns: a) 1818

Soluție:

4⋅abcd‾=dcba‾≤9999⇒a∈{1,2}4 \cdot \overline{abcd} = \overline{dcba} \leq 9999 \textcolor{red}{\Rightarrow} a \in \{1, 2\}

a=1⇒4⋅1bcd‾=dcb1‾a=1 \textcolor{red}{\Rightarrow} 4 \cdot \overline{1bcd} = \overline{dcb1} - nu conv. (par≠imp.)(par \not = imp.)

a=2⇒4⋅2bcd‾=dcb2‾⇒Uc(4⋅d)=2\boxed{a=2} \textcolor{red}{\Rightarrow} 4 \cdot \overline{2bcd} = \overline{dcb2} \textcolor{red}{\Rightarrow} U_c(4 \cdot d) = 2
⇒d∈{3,8}\hspace*{2em} \textcolor{red}{\Rightarrow} d \in \{3, 8 \}
∙d=3\hspace*{2em}\bullet d=3 - nu convine (4⋅2bc3‾>3cb2‾)(4 \cdot \overline{2bc3} \gt \overline{3cb2})
∙d=8⇒4⋅2bc8‾=8cb2‾\hspace*{2em}\bullet \boxed{d=8} \textcolor{red}{\Rightarrow} 4 \cdot \overline{2bc8} = \overline{8cb2}
8000+bc‾⋅40+32=8000+cb‾⋅10+2\hspace*{4em} \cancel{8000} + \overline{bc} \cdot 40 + 32 = \cancel{8000} + \overline{cb} \cdot 10 + 2
40⋅bc‾+30=10⋅cb‾∣:10\hspace*{4em} 40 \cdot \overline{bc} + 30 = 10 \cdot \overline{cb} \quad | :10
4⋅bc‾+3=cb‾\hspace*{4em} 4 \cdot \overline{bc} + 3 = \overline{cb}
\hspace*{4em}Dar cb‾≤99⇒b∈{0,1,2}\overline{cb} \leq 99 \textcolor{red}{\Rightarrow} b \in \{0, 1, 2\}
∙b−par\hspace*{4em}\bullet b - par, nu conv. (par+imp.≠par)(par + imp. \not = par)
∙b=1⇒4(1⋅10+c)+3=c1‾\hspace*{4em}\bullet \boxed{b = 1} \textcolor{red}{\Rightarrow} 4(1 \cdot 10 + c) + 3 = \overline{c1}
43+4c‾=c1‾⇒Uc(4⋅c)=8\hspace*{6em} 43+\overline{4c} = \overline{c1} \textcolor{red}{\Rightarrow} U_c(4 \cdot c)=8
⇒c∈{2,7}\hspace*{6em} \textcolor{red}{\Rightarrow} c \in \{2, 7\}
∙c=2⇒43+4⋅2=21\hspace*{6em}\bullet c=2 \textcolor{red}{\Rightarrow} 43+4 \cdot 2=21 - fals
∙c=7⇒43+4⋅7=71\hspace*{6em}\bullet \boxed{c=7} \textcolor{red}{\Rightarrow} 43+4 \cdot 7=71 - adev.

Deci a+b+c+d=2+1+7+8=18.a+b+c+d = 2+1+7+8 = 18.