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Problem Of The Week 2

with 2 comments


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

Written by Implex

August 27, 2008 at 8:32 am

2 Responses

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  1. 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 sharma

    August 27, 2008 at 10:38 am

  2. very nicely done, thats correct !

    outtimed

    August 27, 2008 at 10:44 am


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