#n# is the smallest positive integer such that (2001+n) is the sum of the cubes of the first 'm' natural numbers. Then (m,n)=?

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Answer 1 · 2018-07-29T06:08:35

Sum of the cubes of the first 'm' natural numbers is

#=((m(m+1))/2)^2=(2001+n)#

So #(2001+n)# must be perfect square. The smallest value of n for which #(2001+n)# is a perfect square, will be #n=24#

Hence #((m(m+1))/2)^2=(2001+24)#

#=>(m(m+1))/2=sqrt(2001+24)=45#

#=>m^2+m-90=0#

#=>m^2+10m-9m-90=0#

#=>m(m+10)-9(m+10)=0#

#=>(m+10)(m-9)=0#

So #m=9# as m is the positive integer.