Question 1017725
{{{N}}}= the original number
{{{N(85/100)}}}= 85% of the original number
{{{N+Nk/100=N((100+k)/100)}}}= the original number increased by {{{"k %"}}} .
Increasing a number by {{{"k %"}}} is multiplying it by {{{((100+k)/100)}}} .
Similarly, decreasing a number {{{S}}} by {{{"w %"}}} is multiplying it by {{{((100-w)/100)}}} :
{{{S-Sw/100=S(1-w/100)=S((100-w)/100)}}} .
So,
{{{N((100+k)/100)((100-w)/100)}}}= the original number first increased by {{{"k %"}}} , and then decreased by {{{"w %"}}} .
Our equation is
{{{N((100+k)/100)((100-w)/100)=N(85/100)}}}
{{{((100+k)/100)((100-w)/100)=(85/100)}}}
{{{((100+k)/100)(100-w)=85}}}
{{{(100+k)(100-w)=85*100}}}
{{{(100+k)(100-w)=8500}}}
Now we need pairs of factors of {{{8500=85*10*10=17*5*2*5*2*5=2^2*5^3*17=2^2*5^3*17^1}}}
There should be {{{(2+1)(3+1)(1+1)=3*4*2=24}}} factors, which come in {{{24/2=12}}} pairs:
{{{1*8500=8500}}} ,
{{{2*4250=8500}}} ,
{{{4*2125=8500}}} ,
{{{5*1700=8500}}} ,
{{{10*850=8500}}} ,
{{{17*500=8500}}} ,
{{{20*425=8500}}} ,
{{{25*340=8500}}} ,
{{{34*250=8500}}} ,
{{{50*170=8500}}} ,
{{{68*125=8500}}} , and
{{{85*100=8500}}} .
Since {{{0<k<173}}} , {{{100<100+k<273}}} .
The only factors that could be {{{100+k}}} are {{{125}}} , {{{170}}} , and {{{250}}} .
{{{100+k=125}}}--->{{{k=125-100}}}--->{{{highlight(k=25)}}} , or
{{{100+k=170}}}--->{{{k=170-100}}}--->{{{highlight(k=70)}}} , or
{{{100+k=250}}}--->{{{k=250-100}}}--->{{{highlight(k=150)}}} .