Question

Micsorans de 3 ori suma a 6 numere consecutive,obtin 21.
Care este penultimul numar?
Prin segmente si intrebari.

Answer 1

a+a+1+a+2+a+3+a+4+a+5=x÷3
x÷3=21
x=63
6a+15=63
6a=48
a=8
a+1=9
a+2=10
a+3=11
a+4=12===>penultimul număr
a+5=13

Answer 2

A |---|
B |---|+1
C |---|+2
D |---|+3
E |---|+4
F |---|+5
SUMA = 21 × 3 = 63
63 - 1 - 2 - 3 - 4 - 5 =
62 - 2 - 3 - 4 - 5 =
60 - 3 - 4 - 5 =
57 - 4 - 5 =
53 - 5 = 48 ( SUMA A ŞASE SEGMENTE EGALE )
A = 48 : 6 = 8
B = 8 + 1
B = 9
C = 8 + 2
C = 10
D = 8 + 3
D = 11
E = 8 + 4
E = 12
F = 8 + 5
F = 13

VERIFICARE :

8 + 9 + 10 + 11 + 12 + 13 =
17 + 10 + 11 + 12 + 13 =
27 + 11 + 12 + 13 =
38 + 12 + 13 =
50 + 13 = 63

63 : 3 = 21