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06 Dec 2018, 00:19
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B
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Difficulty:
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76% (00:54) correct
24%
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What is the units digit of \(3^{23} - 2^{18}\) ?
A. 1
B. 3
C. 5
D. 7
E. 9
A. 1
B. 3
C. 5
D. 7
E. 9
06 Dec 2018, 00:28
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Bunuel
What is the units digit of \(3^{23} - 2^{18}\) ?
A. 1
B. 3
C. 5
D. 7
E. 9
A. 1
B. 3
C. 5
D. 7
E. 9
\(3^{23} - 2^{18}\) ?
\(3^{23}\) = 3 has cyclic of 4 ( unit digit repeats ever 4th power) so the last digit would be same as 23- (4*5) = \(3^{3}\) which will have unit digit 7 ( \(3^{3}\) = 27)
similarly
\(2^{18}\) = 2 has cyclic of 4 as well so the last digit will be same as 18- (4*4) = \(2^{2}\) which will have unit digit 4 ( \(2^{2}\)=4)
so \(3^{23} - 2^{18}\) = 7 - 4 = 3 ans B
anshkool4u
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06 Dec 2018, 04:46
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Answer is B.
Since 3 has a cycle of 4,when raised to the power, and unit digit of 3^4 is 1.This means 3^23 can be written as 3^(20+3)= (3^20)*(3^3).
Since 3^3 is 27, unit digit is 7.
Thus unit digit of 3^23 =(1)(7)= 7
2 has a cycle of 4, 2^18 can be written as (2^16)*(2^2). Unit digit of 2^4 is 6 and unit digit of 2^2 is 4, so unit digit of 2^18 = 6*4=24, unit digit is 4.
Hence unit digit of 3^23-2^18= 7-4= 3
Hence B
Since 3 has a cycle of 4,when raised to the power, and unit digit of 3^4 is 1.This means 3^23 can be written as 3^(20+3)= (3^20)*(3^3).
Since 3^3 is 27, unit digit is 7.
Thus unit digit of 3^23 =(1)(7)= 7
2 has a cycle of 4, 2^18 can be written as (2^16)*(2^2). Unit digit of 2^4 is 6 and unit digit of 2^2 is 4, so unit digit of 2^18 = 6*4=24, unit digit is 4.
Hence unit digit of 3^23-2^18= 7-4= 3
Hence B
