how many area codes are possible if the first digit can’t be zero, no digit can be repeated, and the third digit must be 9 or 0?
1 answer:
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Solution:
The given equation is:
A: Taking , x=1
L H S = 9 × 1 + 49 = 9 + 49= 58
R H S :
L H S ≠ R H S
B: Taking x=3
L H S = 9 × 3 + 49=27 + 49=76
R HS =
L H S ≠ R H S
C: Taking x=5
L H S = 9 × 5 + 49
= 45 + 49
= 94
R H S:
L H S ≠ R H S
(D)
when , x= 8
L H S
9 × 8 + 49
= 72 + 49
= 121
R HS:
L H S ≠ R H S
