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S(n)= 4^n +5^(n+1) +3 what is the unit digit of s(100)

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Joined: 24 May 2010
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S(n)= 4^n +5^(n+1) +3 what is the unit digit of s(100) [#permalink]

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31 Jul 2010, 14:50
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S(n)= 4^n +5^(n+1) +3 what is the unit digit of s(100)
Ms. Big Fat Panda
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Joined: 09 Jun 2010
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Concentration: General Management, Nonprofit
Re: S(n)= 4^n +5^(n+1) +3 what is the unit digit of s(100) [#permalink]

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31 Jul 2010, 15:04
$$S(n)= 4^n +5^(n+1) +3$$

The way to solve this would be to look at each individual term.
$$4^1 = 4$$
$$4^2 = 16$$
$$4^3 = 64$$
and so on. So, the cyclicity of this is 2. Every 2 terms the last digit is 4 when the n is odd and 6 when n is even.

With 5, its always 5.

If n = 100, it's an even number and hence 4^100 will end in 6 and 5^101 will end in 5.

Hence the sum of last digits will be $$6+5+3 = 14$$ so it'd be 4.
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Re: S(n)= 4^n +5^(n+1) +3 what is the unit digit of s(100) [#permalink]

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02 Aug 2010, 07:22
If n = 100, it's an even number and hence 4^100 will end in 6 and 5^101 will end in 5.

6+5+3
; so it'd be 4.
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Re: S(n)= 4^n +5^(n+1) +3 what is the unit digit of s(100) [#permalink]

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18 Apr 2017, 11:40
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Re: S(n)= 4^n +5^(n+1) +3 what is the unit digit of s(100)   [#permalink] 18 Apr 2017, 11:40
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