Question

Aflati numerele abcd şi n pentru care abcd + a + b + c + d = n!
(unde n! = 1 x 2 x 3 ..... x n şi 0! = 1). abcd este scris in baza 10 cu liniuta deasupra. va rog mult ajutor​

Answer 1

$\displaystyle\bf\\\overline{abcd}+a+b+c+d~poate~fi~cel~mult~9999+9+9+9+9=10035,\\si~cel~putin~1001.\\deci,~1001\leq n!\leq 10035 \implies doar~7!~implineste~aceasta~conditie,~\\deci~\boxed{\bf n=7}~.\\\overline{abcd}+a+b+c+d=5040 \Leftrightarrow 1001a+101b+11c+2d=5040,~de~unde\\a~poate~lua~valorile~in~\mathbb{N}~de~la~1~pana~la~5,~dar~daca~a\geq 4 \implies\\1001a+101b+11c+2d\leq 4004+909+99+18 =5030,~deci~\boxed{\bf a=5}~.\\5005+101b+11c+2d=5040 \Leftrightarrow 101b+11c+2d=35.$

$\displaystyle\bf\\de~unde~evident~\boxed{\bf b=0}~.\\11c+2d=35 \Leftrightarrow 11c=35-2d,~cum~35-2d~nu~poate~fi~22,~dar~poate~\\fi~33~\implies \boxed{\bf c=3}~,~\boxed{\bf d=1}~.\\\boxed{\bf \overline{abcd}=5031}$