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30 Nov 2016, 03:00
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E
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Difficulty:
15%
(low)
Question Stats:
86% (01:14) correct
14%
(01:17)
wrong
based on 72
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Let k = 19^19 + 23^23. What is the units digit of k?
A. 0
B. 2
C. 4
D. 5
E. 6
A. 0
B. 2
C. 4
D. 5
E. 6
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30 Nov 2016, 04:13
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To solve this question, we need to worry about the tenth digit of both the number. Unit digit is sufficient to calculate the unit digit of K.
9^1=9 3^1=3
9^2=1 3^2=9
9^3=9 3^3=7
9^4=1 3^3=1
This values will repeat as we go ahead. Hence, unit digit of 19^19 will be 9. 23^23 unit digit will be 7. Hence last digit of k will be 9+7=16 as 6. My answer will be E.
9^1=9 3^1=3
9^2=1 3^2=9
9^3=9 3^3=7
9^4=1 3^3=1
This values will repeat as we go ahead. Hence, unit digit of 19^19 will be 9. 23^23 unit digit will be 7. Hence last digit of k will be 9+7=16 as 6. My answer will be E.
30 Nov 2016, 09:37
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Bunuel
Let k = 19^19 + 23^23. What is the units digit of k?
A. 0
B. 2
C. 4
D. 5
E. 6
A. 0
B. 2
C. 4
D. 5
E. 6
\(9^{Odd}\) = Units digit 9 so, units digit of \(19^{19}\) will be \(9\)
Cyclucity of 3 is 4 , so units digit of \(23^{23}\) will be \(7\)
Hence, units digit of \(19^{19}\) + \(23^{23}\) = 9 + 7 = 16
Answer will be (E)
