Question

Numarul natural n, scris in baza 10, are suma cifrelor egala cu 2009. Determinati suma cifrelor numarului n+7. (Gasiti toate posibilitatile. ).

Answer 1

Răspuns:

2007; 2016

Explicație pas cu pas:

Numarul natural n, scris in baza 10, cu n cifre:

$n = \overline {a_{1}a_{2}a_{3}...a_{n-1}a_{n}}; \ \ a_{1} \not= 0$

$a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2009$

$a_{n} = 0 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2009 \\ 0 + 7 = 7 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2009 + 7 = \bf 2016 \\ a_{n} = 1 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2008 \\ 1 + 7 = 8 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2008 + 8 = \bf 2016 \\ a_{n} = 2 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2007 \\ 2 + 7 = 9 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2007 + 9 = \bf 2016 \\ a_{n} = 3 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2006 \\ 3 + 7 = 10 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2006 + 1 + 0 = \bf 2007 \\ a_{n} = 4 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2005 \\ 4 + 7 = 11 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2005 + 1 + 1 = \bf 2007 \\ a_{n} = 5 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2004 \\ 5 + 7 = 12 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2004 + 1 + 2 = \bf 2007 \\ a_{n} = 6 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2003 \\ 6 + 7 = 13 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2003 + 1 + 3 = \bf 2007 \\ a_{n} = 7 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2002 \\ 7 + 7 = 14 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2002 + 1 + 4 = \bf 2007 \\ a_{n} = 8 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2001 \\ 8 + 7 = 15 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2001 + 1 + 5 = \bf 2007 \\ a_{n} = 9 \implies a_{1}+a_{2}+a_{3}+...+a_{n-1} = 2000 \\ 9 + 7 = 16 \iff a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} = 2000 + 1 + 6 = \bf 2007$

→ există două posibilități: 2007 sau 2017