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05 Sep 2023, 01:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
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Question Stats:
83% (01:29) correct
17%
(01:58)
wrong
based on 89
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What is the unit digit of integer \((3^{77})(7^{33})+(5^{66})(6^{55})\)?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
05 Sep 2023, 04:45
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The units digits of the powers of 3 follow a repetitive pattern of 3,9,7,1,3,9,7,1... So the 77th power would have a units digit of 3.
The units digits of the powers of 7 follow a repetitive pattern of 7,9,3,1,7,9,3,1... So the 33rd power would have a units digit of 7.
So, for the first number, the units digit will be 3+7 = 0
All powers of 5 end with a units digits of 5. All powers of 6 end with a units digit of 6.
So, for the second number, the units digit will be 5+6 = 1
0+1 = 1. So, the units digit of the eventual integer will be 1. Hence, the answer is Option A
The units digits of the powers of 7 follow a repetitive pattern of 7,9,3,1,7,9,3,1... So the 33rd power would have a units digit of 7.
So, for the first number, the units digit will be 3+7 = 0
All powers of 5 end with a units digits of 5. All powers of 6 end with a units digit of 6.
So, for the second number, the units digit will be 5+6 = 1
0+1 = 1. So, the units digit of the eventual integer will be 1. Hence, the answer is Option A
05 Sep 2023, 06:45
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Bunuel
This is based on concept of cyclicity at unit's place.
The unit digit of power of 3 repeats after every 4 powers
e.g. \(3^1 = 3\)
\(3^2 = 9\)
\(3^3 = 27\)
\(3^4 = 81\)
\(3^5 = 243\) Unit digit repeats
Cyclicity of numbers with unit digit {2, 3, 7, 8} is 4
Cyclicity of numbers with unit digit {0, 1, 5, 6} is 1
Cyclicity of numbers with unit digit {4, 9} is 2
The unit digit of power of 3 repeats after every 4 powers
e.g. \(3^1 = 3\)
\(3^2 = 9\)
\(3^3 = 27\)
\(3^4 = 81\)
\(3^5 = 243\) Unit digit repeats
Cyclicity of numbers with unit digit {2, 3, 7, 8} is 4
Cyclicity of numbers with unit digit {0, 1, 5, 6} is 1
Cyclicity of numbers with unit digit {4, 9} is 2
\((3^{77})(7^{33})+(5^{66})(6^{55}) = (3^{4a+1})(7^{4b+1})+(5^{66})(6^{55}) = = 3^1*7^1+5*6 = 3*7+5*6 = 1+0 = 1 \)
Answer: Option A
A good question on related concept:
