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at unit’s place

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by sanju09 » Fri May 14, 2010 1:48 am
What is the last (at unit's place) digit in the product (2^1)(2^2)(2^3)(2^4)...(2^198)(2^199)(2^200)?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
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by kstv » Fri May 14, 2010 7:25 am
2¹*2².............2^200= 2^(1+2........200)
sum of n consecutive nos = n(n+1)/2

2¹=2,2²=4,8,16,32,64,128,256
the pattern of the units place is
2 4 8 6
same as the answer options

IMO D
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by indiantiger » Fri May 14, 2010 8:02 am
Is there a quick way to do these type of questions? I can get the answer but takes me a lot of time.
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by analyst218 » Fri May 14, 2010 8:49 am
indiantiger wrote:Is there a quick way to do these type of questions? I can get the answer but takes me a lot of time.
well,

all you need to do is find the power cycle, as soon as you realize its 2 4 8 6 , then you immediately get to the answer since 200 is a multiple of 4. it shouldn't take you more than a min
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by gmatjedi » Fri May 14, 2010 2:28 pm
first set up:
2^1=2
2^2=4
2^3=8
2^4=16
cycle then repeats

realize that the lower powers can be paired with the higher powers and will sum to an exponent of 200
ex. 2^1*2^199=2^200
ex. 2^2*2^198=2^200
ex. 2^3*2^197=2^200
and so on...

therefore based on the cycle above 2^200 should end terminate with 6
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