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PS: Exponents (m06q07) [#permalink]
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13 Nov 2008, 11:25
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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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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Re: PS: Exponents (m06q07) [#permalink]
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Re: PS: Exponents (m06q07) [#permalink]
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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 [#permalink]
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13 Nov 2008, 12:38
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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]
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16 Sep 2009, 19:45
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2^98=256L + N 2^98=2^8L + N 2^982^8L = N N=2^8(2^90L) Taking L=2^90,N=0
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Re: PS: Exponents (m06q07) [#permalink]
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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 [#permalink]
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13 Nov 2008, 12:32
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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Re: PS: Exponents (m06q07) [#permalink]
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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]
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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]
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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]
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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]
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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]
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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]
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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 , 2^80/2^4=256/16=16.000 to n = 2 , 2^82/2^4=2562/16=254/16=15.875 Bests, Cesar
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Re: PS: Exponents (m06q07) [#permalink]
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04 Apr 2012, 06:44
Bunuel, awesome answer =)
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Re: PS: Exponents (m06q07) [#permalink]
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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]
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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]
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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]
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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]
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05 Apr 2013, 05:26
N = 0 makes LHS = 2^98 RHS = 2^8 * 2^90 + 0
IMO A




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