 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Fraction and Root (Posted on 2010-09-30) 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]

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) analytical solution Comment 1 of 1

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 = (88-v)/100 we need
v^2/100 <= (88-v)/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^2-v<=186
v(v-1)<=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 2010-09-30 23:55:15 Please log in:

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