vikrambansal wrote:Unit's digit of 33^43 = unit's digit of (4*8 + 1)^43 = 3
Unit's digit of 43^33 = unit's digit of (4*10 + 3)^43 = 7
If 43 = 4*10 + 3 then why not 33 = 3*10 + 3? Then the answer would be 7 + 7 = 14 or 4 (C)
3¹ --> units digit of 3.
3² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)
3³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)
3� --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)
From here, the units digits will repeat in the same pattern: 3, 9, 7, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus:
33�� and 43³² each have a units digit of 1.
From here, the cycle of units digits will repeat: 3, 9, 7, 1...
Thus:
33�¹ and
43³³ each have a units digit of 3.
33�² has a units digit of 9.
33�³ has a units digit of 7.
Result:
Since n =
33�³ +
43³³, n ---> (units digit of 3) + (units digit of 7) = units digit of 0.
The correct answer is
A.
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