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PS: Exponents (m06q07) [#permalink]
 13 Nov 2008, 11:25
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If 2^{98} = 256L + N , where L and N are integers and 0 \le N \le 4 , what is the value of N ? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Source: GMAT Club Tests - hardest GMAT questions
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bigfernhead wrote: If 2^(98) = 256L + Nwhere L and N are integers and 0<=N<=4, what is the value of N?
0
1
2
3
4 2^(98) = (2^8) L + N N should be 0 otherwise L wont be an integer.
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bigfernhead wrote: If 2^(98) = 256L + Nwhere L and N are integers and 0<=N<=4, what is the value of N?
0
1
2
3
4 2^98=256L + N=2^8L + N considering 0<=N<=4 N has to be 0 and L=2^90 IMO A
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Re: PS: Exponents (m06q07) [#permalink]
16 Sep 2009, 19:45
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2^98=256L + N 2^98=2^8L + N 2^98-2^8L = N N=2^8(2^90-L) Taking L=2^90,N=0
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Re: PS: Exponents (m06q07) [#permalink]
29 Mar 2010, 05:02
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Answer A
As I couldn't find a way to solve this, I tested each solution
A. N=0
2^98=2^8*L, hence L=2^90 which is possible
B. N=1
2^98=2^8*L+1, 2^98 is even and 2^8*L+1 is odd so impossible
C. N=2
2^98=2^8*L+2
<=>2^97=2^7+1, 2^97 is even and 2^7*L+1 is odd so impossible
At this point in real test I would have chosen N=0 because N=3 is the "same" as N=1 and N=4 is the "same" as N=2
but , in order to finish, here is the rest :
D. N=3
2^98=2^8*L+3, 2^98 is even and 2^8*L+3 is odd so impossible
E. N=4
2^98=2^8*L=4
<=>2^96=2^6*L+1, 2^96 is even and 2^6*L+1 is odd so impossible
Answer A
My question is simple : is there a better way to solve this question ?
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Re: PS: Exponents (m06q07) [#permalink]
29 Mar 2010, 08:57
2^98=2^8L + N
Since, 0<=N<=4 , the only feasible solution is N=0
Answer: A
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Re: PS: Exponents (m06q07) [#permalink]
29 Mar 2010, 09:20
last digit of 2*98 is 4 ( 2*98 = 2*2 )
this happen when
a) N = 0 & L = 2*90
b) N = 2 & L = 2*89
Who is not obedient, I'll give details
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Re: PS: Exponents (m06q07) [#permalink]
29 Mar 2010, 14:56
ykpgal wrote: last digit of 2*98 is 4 ( 2*98 = 2*2 )
this happen when
a) N = 0 & L = 2*90
b) N = 2 & L = 2*89
Who is not obedient, I'll give details I am confused. How did you get N=2? Because N=2 reduces the given equation to 2^98= 2^8 * L + 2 => 2^98 = 2 (2^7*L + 1) => 2^97 = 2^7 L +1 -> Incorrect, as Right side is odd and Left side is even.
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Re: PS: Exponents (m06q07) [#permalink]
29 Mar 2010, 22:11
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2^{98} = 256L + N
The given expression is a classic example of division:
dividend (2^{98}) = quotient (L)*divisor (256=2^8) + remainder (N).
Since 2^{98}/2^8=2^{90}=L the remainder N=0.
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Re: PS: Exponents (m06q07) [#permalink]
30 Mar 2010, 01:28
I saw it today ( my country time is CET )
the number inside brackets 1 not represent N
The last digit of 2*98 is 4 ( I have sent a theory about last digit few months ago )
This happen
a) when 256∙L have last digit 4 and N=0 ( ..6 ∙ ..4 give 4, this happen L=2*90,=2*2)
b) when 256∙L have last digit 2 and N=2 (...6 ∙ ..2 give 2, this happen L=2*89 =2*1)
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Re: PS: Exponents (m06q07) [#permalink]
31 Mar 2011, 06:06
2^98 = 256L + N => 2^98 = 2^8 * L + 4 OR 2^8 * L + 2 OR 2^8 * L + 0 2^98 = 2^8 * 2^90 + 0 => N = 0 Answer - A
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Re: PS: Exponents (m06q07) [#permalink]
06 Apr 2011, 13:08
I tried to compare it with a similar example of less power
like 2^6 = 2^3L + N
and reached the conclusion that it can only be 0.
though I liked nvgroshar's reasoning better !
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Re: PS: Exponents (m06q07) [#permalink]
04 Apr 2012, 06:07
Hi, Please correct me if I´m wrong. In options 2 and 4, the number is a FRACTION, and not a odd/even division, Like L = 2^98 - 2 / 2^8 will run from 2^n/2^m exponencial divison so will be a fraction. Example = 2^8 - n / 2^4 to n = 0 , 2^8-0/2^4=256/16=16.000 to n = 2 , 2^8-2/2^4=256-2/16=254/16=15.875 Bests, Cesar
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Re: PS: Exponents (m06q07) [#permalink]
04 Apr 2012, 06:33
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Re: PS: Exponents (m06q07) [#permalink]
04 Apr 2012, 06:44
Bunuel, awesome answer =)
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Re: PS: Exponents (m06q07) [#permalink]
04 Apr 2012, 07:07
bigfernhead wrote: If 2^{98} = 256L + N , where L and N are integers and 0 \le N \le 4 , what is the value of N ? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Source: GMAT Club Tests - hardest GMAT questions Answer is A Reason why given 2^98 = 2^8 * L + N if both sides have to be integers and if 2^8 * L have to be integer, we get it divide 2^98 and it divides by 2^90.... Hence N = 0
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Re: PS: Exponents (m06q07) [#permalink]
04 Apr 2012, 15:18
I was able to deduce the part where 256 = 2^90 and then get the 2^8 but I did not realize that this equation was the "dividend=quotient * divisor + remainder." This was a tricky question for sure and I guess now after seeing the solution I understand why A=0. I'll have to put this in my error log so I know why in the future
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Re: PS: Exponents (m06q07) [#permalink]
04 Apr 2012, 21:15
IMO N=0
because
2^8 t0 2^98 is very big jump..... to get 2^98 from 2^8 we need to ad very big number but here we can ad at max 4...
hence n has to be zero
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Re: PS: Exponents (m06q07) [#permalink]
04 Apr 2012, 22:30
nvgroshar wrote: 2^{98} = 256L + N
The given expression is a classic example of division:
dividend (2^{98}) = quotient (L)*divisor (256=2^8) + remainder (N).
Since 2^{98}/2^8=2^{90}=L the remainder N=0. Lovely way to arrive at the answer 
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Re: PS: Exponents (m06q07) [#permalink]
05 Apr 2013, 05:26
N = 0 makes
LHS = 2^98
RHS = 2^8 * 2^90 + 0
IMO A
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Re: PS: Exponents (m06q07)
[#permalink]
05 Apr 2013, 05:26
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