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# PS: Exponents (m06q07)

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PS: Exponents (m06q07) [#permalink]

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13 Nov 2008, 10:25
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This topic is locked. If you want to discuss this question please re-post 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

[Reveal] Spoiler: OA
A

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Re: PS: Exponents [#permalink]

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13 Nov 2008, 11:32
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 [#permalink]

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13 Nov 2008, 11:38
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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, 18:45
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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]

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29 Mar 2010, 04:02
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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]

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29 Mar 2010, 07:57
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, 08:20
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, 13:56
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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29 Mar 2010, 21:11
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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]

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30 Mar 2010, 00:28
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, 05:06
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, 12:08
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, 05:07
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]

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04 Apr 2012, 05:33
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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

[Reveal] Spoiler: OA
A

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Given: $$2^{98}=2^8*L+N$$ --> divide both parts by $$2^8$$ --> $$2^{90}=L+\frac{N}{2^8}$$. Now, as both $$2^{90}$$ and L are an integers then $$\frac{N}{2^8}$$ must also be an integer, which is only possible for $$N=0$$ (since $$0\leq{N}\leq4$$).

Answer: A.
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Re: PS: Exponents (m06q07) [#permalink]

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04 Apr 2012, 05:44
Bunuel,
awesome answer
=)
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Re: PS: Exponents (m06q07) [#permalink]

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04 Apr 2012, 06:07
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

[Reveal] Spoiler: OA
A

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, 14:18
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, 20:15
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, 21:30
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, 04:26
N = 0 makes
LHS = 2^98
RHS = 2^8 * 2^90 + 0

IMO A
Re: PS: Exponents (m06q07)   [#permalink] 05 Apr 2013, 04:26

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# PS: Exponents (m06q07)

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