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12 Jun 2024, 01:31
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
77% (01:02) correct
23%
(01:12)
wrong
based on 229
sessions
History
Date
Time
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What is the units digit of \((23^6)(17^3)(61^9)\)?
A. 1
B. 3
C. 5
D. 7
E. 9
A. 1
B. 3
C. 5
D. 7
E. 9
12 Jun 2024, 02:28
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I think its D, 7.
(23^6)(17^3)(61^9)
to calculate unit digit of above values we need to check unit digit of each bracket.
23^6 in the form of "_________9" where 9 is unit digit calculated by using cyclicity of digit 3.
17^3 in the form of "_________3" where 3 is unit digit again using cyclicity of digit 7
61^9 in the form of "_________1" where 1 is unit digit cyclicty of digit 1.
Now unit digit (23^6)(17^3)(61^9) = "______9" * "_______3" * "_______1" = 9*3*1= "_____7"
(23^6)(17^3)(61^9)
to calculate unit digit of above values we need to check unit digit of each bracket.
23^6 in the form of "_________9" where 9 is unit digit calculated by using cyclicity of digit 3.
17^3 in the form of "_________3" where 3 is unit digit again using cyclicity of digit 7
61^9 in the form of "_________1" where 1 is unit digit cyclicty of digit 1.
Now unit digit (23^6)(17^3)(61^9) = "______9" * "_______3" * "_______1" = 9*3*1= "_____7"
Shwarma
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12 Jun 2024, 03:37
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Option D
(_3)^ 6 = 3 power works in mutliples of 4
so 3^6 = 3^2 = last digit is 9
(_7)^3 - 7^3 unit digit is 3
1^n is always 1 so ignore
so unit digit is : (9*3*1) = _7
So option D= 7
(_3)^ 6 = 3 power works in mutliples of 4
so 3^6 = 3^2 = last digit is 9
(_7)^3 - 7^3 unit digit is 3
1^n is always 1 so ignore
so unit digit is : (9*3*1) = _7
So option D= 7
