Question

Dau coroana și ce mai vreți voi! dar va rog sa faceți ce știți de aici, nu trebuie neapărat toate dar ce puteți.​

Answer 1

 

23)

Notatii:

Suma numerelor Naturale Impare de 3 Cifre = SNI3C

Suma numerelor Naturale Pare de 2 Cifre = SNP2C

Rezolvare:

$\displaystyle\bf\\SNI3C=101+103+105+107+...+999\\Numarul~termenilor~(n) este:\\\\n=\frac{999-101}{2}+1=\frac{898}{2}+1=449+1=450~de~numere.\\\\SNI3C=\frac{n(999+101)}{2}=\frac{450\times1100}{2}=450\times550=248400\\\\-----\\ SNP2C=10+12+14+16+...+98\\Numarul~termenilor~(n) este:\\\\n=\frac{98-10}{2}+1=\frac{88}{2}+1=44+1=45~de~numere.\\\\SNP2C=\frac{n(98+10)}{2}=\frac{45\times108}{2}=45\times54=2430\\\\Calculam~diferenta:\\SNI3C-SNP2C=248400-2430=\boxed{\bf245970}$

24)

Trei solutii:

                    S1:                    S2:                    S3:

DO I -           928 -               968 -                 564 -                

UNU             464                 484                   282

UNU             464                 484                   282

25)

$\displaystyle\bf\\\overline{abcd}-\overline{abc}-\overline{ab}-a=2020\\\\(1000a+100b+10c+d)-(100a+10b+c)-(10a+b)-a=2020\\\\1000a+100b+10c+d-100a-10b-c-10a-b-a=2020\\\\1000a-100a-10a-a+100b-10b-b+10c-c+d=2020\\\\a(1000-100-10-1)+b(100-10-1)+c(10-1)+d=2020\\\\889a+89b+9c+d=2020\\\\a\neq0~din~enunt\\a~nu~poate~fi~3~deoarece~~889\times3>2020\\a=1~este~prea~putin~chiar~daca~b=9\\\\\implies~a=2\\\\889\times2=1778\\\\b=2Daca~b=3~atunci 889\times2+89\times3=1778+267=2045>2020\\\\\implies~b=2$

.

$\displaystyle\bf\\889\times2+89\times2=1778+178=1956\\\\d~este~un~numar~de~o~cifra.\\\\\implies~Il~alegem~pe~c~astfel~incat~sa~ramana~pana~la~2020\\o~diferenta~de~maxim~o~cifra.\\\\Daca d=6~atunci:\\889\times2+89\times2+9\times6=1778+178+54=2010\\\\2020-2010=10~prea~mult\\\\Daca~d=7~atunci:\\889\times2+89\times2+9\times7=1778+178+63=2019\\\\2020-2019=1\\\\\implies~c=7~~si~~d=1}\\\\889\times2+89\times2+9\times7+1=1778+178+63+1=2020\\\\Solutia:\\\\\boxed{\bf a=2;~b=2;~c=7;~d=1}$