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01 Oct 2018, 03:35
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
54% (01:34) correct
46%
(01:26)
wrong
based on 155
sessions
History
Date
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Not Attempted Yet
Find out the unit digit of \(x\), if \(x = 1!^1 + 2!^2 + 3!^3 + 4!^4 + 5!^5+......10!^{10}\)
A) 1
B) 3
C) 5
D) 7
E) 9
https://www.GMATinsight.com
A) 1
B) 3
C) 5
D) 7
E) 9
https://www.GMATinsight.com
01 Oct 2018, 04:30
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The one thing we need to know before attempting this question is that the last digit of any factorial starting from 5 and beyond is always zero.
So, effectively we need to find the last digits of the starting terms, that is 1!^1, 2!^2, 3!^3, 4!^4,
1!^1 = last digit 1
2!^2 = last digit 4
3!^3 => 3! is 6 (cyclicity of 6 is 1) so last digit is 6
4!^4 => 4! is 24 (cyclicity of last digit of 24 that is 4 is 2) first is 4 second is 6 so, last digit is 6
So adding all
1+4+6+6 = 17 => last digit is 7
So, effectively we need to find the last digits of the starting terms, that is 1!^1, 2!^2, 3!^3, 4!^4,
1!^1 = last digit 1
2!^2 = last digit 4
3!^3 => 3! is 6 (cyclicity of 6 is 1) so last digit is 6
4!^4 => 4! is 24 (cyclicity of last digit of 24 that is 4 is 2) first is 4 second is 6 so, last digit is 6
So adding all
1+4+6+6 = 17 => last digit is 7
General Discussion
01 Oct 2018, 04:26
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X = 1 + 2^2 + 6^3 + 24^4 + 120^5 + 720^6....
X = 1+4+..6+..6+..0+..0+....
X=5+..2
X=..7
If the explaination helped please give kudos.
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X = 1+4+..6+..6+..0+..0+....
X=5+..2
X=..7
If the explaination helped please give kudos.
Posted from my mobile device
