Question

Determinati numerele prime a <b<c<d, stiind că
79 x a + 90 x b +21 x c +77 x d = 2020.
va rog muuult sa ma ajutati! ​

Answer 1

$\displaystyle\bf\\79a+90b+21c+77d=2020,~a,b,c,d~prime,~a<b<c<d.\\in~primul~rand~analizam~paritatea,~si~stim~ca~doar~a~poate~fi~2,\\a<b<c<d,~minim~2<3<5<7,~deci~vedem~daca~a~este~2.\\79a+90b+21c+77d=par,~90b=par,~21c=impar,~77d=impar.\\79a+(par+impar+impar)=par \Leftrightarrow 79a+par=par \implies 79a=par,~deci~\boxed{\bf a=2}~.\\158+90b+21c+77d=2020\Leftrightarrow 90b+21c+77d=1862.\\90b+7(3c+11d)=1862,~dar~1862=\mathcal{M}_7 \implies cum~90b+\mathcal{M}_7=\mathcal{M}_7 \implies$

$\displaystyle\bf\\90b=\mathcal{M}_7,~cum~90~nu~este~divizibil~cu~7\implies b=\mathcal{M}_7,~dar~b=prim,\\singurul~numar~prim~divizibil~cu~7~este~7,~deci~\boxed{\bf b=7}~.\\630+7(3c+11d)=1862 \Leftrightarrow 7(3c+11d)=1232 \Leftrightarrow 3c+11d=176,~\\observam~ca~176=\mathcal{M}_{11},~de~unde~ne~rezulta~ca~3c+11d=\mathcal{M}_{11} \Leftrightarrow 3c=\mathcal{M}_{11},~cum~3~nu~divide~pe~11,~iar~c~este~prim~iar~singurul~numar~prim~\\\mathcal{M}_{11}~este~11\implies \boxed{\bf c=11}~.$

$\displaystyle\bf\\inlocuim,~33+11d=176 \implies \boxed{\bf d=13}~.$