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Algebra

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3 Mar 2008 08:56:38 IST

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Let f(n) be the least number of times you have to hit the

key in the calculator to get a number less than 2 starting from n, n being a natural number. e.g. f(2) =1and f(5) =2. For how many natural numbers m such that 1<m<2008, is f(m) odd.

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Comments (10)

ERAGON007

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3 Mar 2008 11:24:04 IST

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no.s between 8(( 2² )²) & 256(((2²)²)²)

ie 256 - 8 +1 no.s
=249 numbers

is it correct ??

Hari Shankar

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3 Mar 2008 11:37:29 IST

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no. pls try again

anchit saini

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3 Mar 2008 11:50:10 IST

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242??
sir is it right??

ERAGON007

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3 Mar 2008 11:51:07 IST

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if we create a prog in c++ and tell it print all no.s from 1---2008 with when
continuously ^{[ ]}

till we get a num<2
the smallest num is 8 & largest 256 ( f(n)==3 )
& oh! i forgot to include 1,2,3 (f(n) ==1 )

anchit saini

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3 Mar 2008 11:52:08 IST

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eragon i am getting the answer
as
242

anchit saini

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3 Mar 2008 12:00:10 IST

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i did it like this

f(2)=f(3)=1
f(4)=f(5)=...................f(15)=2
f(16)=.........................f(255)=3
thereafter
till 65535 , ans is 4

so numbers come to be
256-16+2=242

mudit parnami

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3 Mar 2008 12:10:19 IST

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i suppose answer is 242 this is how i did it
sq root of 4 is 2
2 time sq root of 16 is 2
3 time sq root of 256 is 2
now calculate nos lying from 2 to 4 exclude 4
then calculate nos lying from 16 to 256 exclude 256
rate if i am correct

Hari Shankar

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3 Mar 2008 12:36:50 IST

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Anybody can back up his answer with a some math logic. You may have got the answer, put presenting a water-tight method will just improve the way you think.

I'm looking for an equation which determines that n. Gimme the eqn and I'll give ya ur rate

sreeraman nagasubramaniyan

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3 Mar 2008 13:16:13 IST

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general formula:

= (2² - 2) + ( (4²)² - 4²) + ( (16²)⁴ - (16²)²) +....

=> (2²-2) + (2⁸-2⁴) + (2³²- 2¹⁶) + ...

=> -2+2²-2⁴+2⁸-2¹⁶+...till as long as 2ⁿ is within the range

in the case given

2⁸<2008 but2¹⁶ not less than 2008

so we take upto 2⁸

Hari Shankar

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3 Mar 2008 13:39:41 IST

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The answer is 242 as almost all of you have got.

Only I wish you show your approach, so that someone who sees it later doesnt think its pulled out of thin air.

Here, we must have (m)^{1/2^n}

2 or 2ⁿ

log₂m

Then it is easy to see for

3 f(m) = 1

15 f(m) = 2

255 f(m) = 3

256

2¹⁶-1 f(m)= 4

That's how you get sboosy's general formula and the answer as 242.

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