Question

Determinati numarul natural abc cu proprietatea ca abc=b la b - b - c​

Answer 1

$\displaystyle\it\\\overline{abc}=b^b-b-c,~a,b,c~cifre.\\evident,~100\leq \overline{abc}\leq 999 \implies 100\leq b^b-b-c\leq 999 \implies b\leq 4.\\daca~b\leq 3 \implies b^b-b-c\leq 100,~deci~evident~\boxed{\it b=4}~.\\\overline{abc}=b^b-b-c \Leftrightarrow 100a+10b+c=b^b-b-c \Leftrightarrow 100a+2c=b^b-b-10b,~\\dar~b=4 \implies 100a+2c=4^4-4-40=212,~impartim~membrii~ecuatiei~la~2.\\50a+c=106,~evident~a~va~fi~2,~atlfel~50a+c\leq 100.\\100+c=106 \implies c=6.\\asadar,~\boxed{\it \overline{abc}=246}~.$

Answer 2

Răspuns:

multumesc colegului pseudoecho, solutia lui buna m-a ajutat sa vad unde gresisem eu pusesem un b in loc de 10b , la abc numar

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abc numar=246

Explicație pas cu pas:

100a+10b+c=b^b-b-c

100a+11b+2c=b^b

b  cifra, pt b=1, b^b o cifra,

pt b=2  , 2²=4  .... 1 cifra

b=3, 3³=27 ....2 cifre

pt b4.4^4=256 trei cifre

b=5, 5^5=3125, 4 cifre

convine doar b=4

fie b=4

100a+11b+2c=4^4=256

la bunul simt a=2

11*4+2c=56

44+2c=56

2c=12

c=6

abc numar=246