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02 Nov 2020, 02:00
Kudos
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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:
35%
(medium)
Question Stats:
71% (01:34) correct
29%
(01:16)
wrong
based on 70
sessions
History
Date
Time
Result
Not Attempted Yet
What is the units digit of \(31467^{32}*97645^{23}*(32168^5 + 8652)^{479}\) ?
(A) 0
(B) 1
(C) 2
(D) 5
(E) 8
(A) 0
(B) 1
(C) 2
(D) 5
(E) 8
02 Nov 2020, 02:46
Kudos
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Explanation:
31467^32 ∗ 97645^23 ∗ (32168^5 + 8652)^479
The cyclicity of 7 is 4: 7,9,3,1, so 7^32 is 1
The cyclicity of 5 is 1: 5 , so 5^23 is 5
The cyclicity of 8 is 4: 8,4,2,6 , so 8^5 is 8
1 ∗ 5 ∗ ( __8 + 8652)^479
5 ∗ (____60)^479
5 ∗ 0
0
IMO-A
31467^32 ∗ 97645^23 ∗ (32168^5 + 8652)^479
The cyclicity of 7 is 4: 7,9,3,1, so 7^32 is 1
The cyclicity of 5 is 1: 5 , so 5^23 is 5
The cyclicity of 8 is 4: 8,4,2,6 , so 8^5 is 8
1 ∗ 5 ∗ ( __8 + 8652)^479
5 ∗ (____60)^479
5 ∗ 0
0
IMO-A
02 Nov 2020, 02:49
Kudos
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Bunuel
\((32168^5 + 8652)^{479}\) = even + even = an even number.
\((97645)^{23}\) will always end with a 5 as the cyclicity of 5 is 1.
5 multiplied by an even number will always end with a 0 and therefore the last digit is always 0.
Option A
Arun Kumar
