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03 Jan 2022, 03:15
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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:
45%
(medium)
Question Stats:
64% (01:27) correct
36%
(01:30)
wrong
based on 391
sessions
History
Date
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Not Attempted Yet
What are the last two digits of the number 7^45?
A. 07
B. 27
C. 23
D. 43
E. 49
A. 07
B. 27
C. 23
D. 43
E. 49
03 Jan 2022, 03:30
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7, when rised to its powers, show a recurrent pattern of having 7, 9, 3, 1, 7 as its unit digit. In tens place, it has 0, 4, 4, 0, 0 as recurrent repetition.
So, for any 5th selection as is the case with the exponent 45, last two digits must be 07.
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So, for any 5th selection as is the case with the exponent 45, last two digits must be 07.
Posted from my mobile device
07 Nov 2024, 06:21
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7^45/100 will give the last 2 digits.
45 = 44+1 = 4x11+1
7^(4x11+1) = [7^(4x11)]*7
Cyclicity of 7 runs on powers of 4; 4x11 will give 0 remainder; only contributor to remainder will be a 7.
Therefore the asked last 2 digits will be 07 (Option A)
45 = 44+1 = 4x11+1
7^(4x11+1) = [7^(4x11)]*7
Cyclicity of 7 runs on powers of 4; 4x11 will give 0 remainder; only contributor to remainder will be a 7.
Therefore the asked last 2 digits will be 07 (Option A)
