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Re: PS: Exponents (m06q07) [#permalink]
29 Mar 2010, 21:11
5
This post received KUDOS
\(2^{98} = 256L + N\) 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 \(N=0\).
Re: PS: Exponents (m06q07) [#permalink]
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)
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 !
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
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 \(0\leq{N}\leq4\)).
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
\(2^{98} = 256L + N\) 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 _________________