Determine all possible value(s) of a real number N that satisfy both the following conditions:
100*{N} + [√(100*{N})] = 88, and:
[N]  [√[N]] = 186
Note: [x] is defined as the greatest integer ≤ x, and {x} = x  [x]
let [N]=a and {N}=r, I will solve for a and r seperately
solving for r:
from 100*{N} + [Sqrt(100*{N})] = 88 we have
100r + [10*sqrt(r)] = 88 so
r = (88[10*sqrt(r)])/100
now since 0<=r<1 then we have 0<=[10*sqrt(r)]<10
thus if [10*sqrt(r)] = v with 0<=v<=9 then we have
v<=10*sqrt(r) < v+1
v^2 <= 100r < (v+1)^2
v^2/100 <= r < (v+1)^2/100
since r = (88v)/100 we need
v^2/100 <= (88v)/100 < (v+1)^2/100
quickly checking for all integer v between 0 and 9, we find that this is true only for v=8 thus
r = (88  8) / 100 = 80/100 = 4/5 = 0.8
solving for a:
from
[N]  [sqrt([N])] = 186 we have
a  [sqrt(a)] = 186
now if v=[sqrt(a)] then we have
v<=sqrt(a)<v+1
v^2<=a<(v+1)^2
since a = 186+v we have
v^2<=186+v<(v+1)^2
v^2<=186+v
v^2v<=186
v(v1)<=186
gives v<=14
checking all 15 possible values of 0<=v<=14 the only valid value is v=14
which gives
a= 186+14 = 200
thus N = a+r = 200.8
checking that this solution is valid
100*0.8 + [10*sqrt(0.8)]
80+8
88
and
200  [sqrt(200)]
200  14
186
and the solution checks out

Posted by Daniel
on 20100930 23:55:15 