## Problem Of The Week 2

**Let us define [x] as the greatest integer which is less than or equal to x and logp(q) is logarithm of q to the base p. Find a natural n such that**

**[log2(1)]+[log2(2)]+[log2(3)]+….+[log2(n)]=2008**

**a) 312 b) 313 c) 314 d) 315 e) none of these**

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1,2,4,8,16,32,64,128,256 are powers of 2 that we will encounter here…

no. of terms that give values that are integral, is nothing but the no. of terms between two consecutive powers of 2,meaning:

no.s from 1-2: give log2() as 0(not including last no.)

from 2-4: 1, 2 no.s-2,3

from 4-8: 2, 4 no.s–4,5,6,7

from 8-16: 3, 8 no.s:8,9,10,11,12,13,14,15

16-32: 4, 16 no.s

32-64: 5, 32 no.s, and from 64-128: 6, 64 no.s, and 128-256: 7, 128 no.s…

total is: 2+8+24+64+160+384+896=1538..so no. left is=2008-1538=470, which has 58 8’s and a 6…this means 58 no.s after 256,i.e 313…any other no. will add 8, and sum exceeds 2008

so none of these should be the answer

sumit sharmaAugust 27, 2008 at 10:38 am

very nicely done, thats correct !

outtimedAugust 27, 2008 at 10:44 am