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# Find out the unit digit of x, if x = 1!^1 + 2!^2 + 3!^3 + 4!^4 + 5!^5+

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Find out the unit digit of x, if x = 1!^1 + 2!^2 + 3!^3 + 4!^4 + 5!^5+  [#permalink]

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01 Oct 2018, 02:35
00:00

Difficulty:

55% (hard)

Question Stats:

57% (01:12) correct 43% (01:39) wrong based on 46 sessions

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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

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Re: Find out the unit digit of x, if x = 1!^1 + 2!^2 + 3!^3 + 4!^4 + 5!^5+  [#permalink]

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01 Oct 2018, 03:26
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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Re: Find out the unit digit of x, if x = 1!^1 + 2!^2 + 3!^3 + 4!^4 + 5!^5+  [#permalink]

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01 Oct 2018, 03:30
3
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
1+4+6+6 = 17 => last digit is 7
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Find out the unit digit of x, if x = 1!^1 + 2!^2 + 3!^3 + 4!^4 + 5!^5+  [#permalink]

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08 Oct 2018, 13:15
GMATinsight wrote:
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

http://www.GMATinsight.com

From 5! the unit digit of each such one will be 0.
So we need only the first four.
$$1!^1$$ unit digit is 1
$$2!^2$$ ---> 4
$$3!^3$$ ---> 6
$$4!^4$$ ---> 6
1+4+6+6=17 --->7
Find out the unit digit of x, if x = 1!^1 + 2!^2 + 3!^3 + 4!^4 + 5!^5+ &nbs [#permalink] 08 Oct 2018, 13:15
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