Question

a) Suma a cinci nr consecutive este 105.Determinati nr.



b)Suma a trei nr consecutive este 168.Determinati nr.


c)Suma a doua nr este 47 iar suma rasturnatelor este 173.Determinati cele doua nr.

Answer 1

a+(a+1)+(a+2)+(a+3)+(a+4)=105

5a + 10=105
5a=95
a=95/5=19

19+20+21+22+23=105
si asa si la celelalte
c
ab+cd=47
ba+dc=173

deci
 10a+b+10c+d=47 => 10(a+c)+(b+d)=47 => b+d=47- 10(a+c)
10b+a+10d+c=173 => (a+c)+10(b+d)=173

=> a+c+10[47-10(a+c)]=173
a+c+470-100a-100c=173
297=99a+99c
a+c=3, dar a,d≠0, deci a=1 si c=2
b+d=47-30=17 , deci b=8 si d=9

Verificare:
18+29=47
81+92=173!

Answer 2

a)a+a+1+a+2+a+3+a+4=105
5a+10=105
5a=95
a=19
R:19,20,21,22,23
b)a+a+1+a+2=168
3a+3=168
3a=165
a=55
R:55,56,57