Problem:
The number a=qpβ, where p and q are relatively prime positive integers, has the property that the sum of all real numbers x satisfying
βxββ {x}=aβ x2
is 420, where βxβ denotes the greatest integer less than or equal to x and {x}=xββxβ denotes the fractional part of x. What is p+q?
Answer Choices:
A. 245
B. 593
C. 929
D. 1331
E. 1332 Solution:
Let w=βxβ and f={x} denote the whole part and the fractional part of x, respectively, for which 0β€f<1 and x=w+f.
We rewrite the given equation as
wβ f=aβ (w+f)2
Since aβ (w+f)2β₯0, it follows that wβ fβ₯0, from which wβ₯0.
We expand and rearrange (1) as
af2+(2aβ1)wf+aw2=0
which is a quadratic with either f or w.
For simplicity purposes, we will treat w as some fixed nonnegative integer so that (2) is a quadratic with f. By the Quadratic Formula, we have
f=w(2a1β2aΒ±1β4aββ)
If w=0, then f=0. We get x=w+f=0, which does not affect the sum of the solutions. Therefore, we consider the case for wβ₯1 :
Recall that 0β€f<1, so 2a1β2aΒ±1β4aβββ₯0. From the discriminant, we require that 0β€1β4a<1, or
0<aβ€41β
We consider each part of 0β€f<1 separately:
fβ₯0 From (2), note that a>0,(2aβ1)w<0, and aw2>0. By Descartes' Rule of Signs, we deduce that (2) must have two positive roots, so fβ₯0 is always valid.
Alternatively, from (3) and (4), note that all values of a for which 0<aβ€41β satisfy 1β2a>1β4aβ. We deduce that both roots in (3) must be positive, so fβ₯0 is always valid.
f<1 We rewrite (3) as
f=w(2a1ββ1Β±2a1β4aββ)
From (4), it follows that 2a1ββ₯1/21β=2. The larger root is
As 1+k<1+W1β, we conclude that 1+k is slightly above 1 so that 2W(W+1)β is slightly below 420 , or W(W+1) is slightly below 840 . By observations, we get W=28. Substituting this into ( β ) produces k=291β, which satisfies W+11ββ€k<W1β, as required.
Finally, we solve for a in k=2a1ββ1β2a1β4aββ :
Since a>0, we obtain 292602βaβ294β=0, from which
a=294ββ 602292β=90029β
The answer is 29+900=(C)929β .
OR
Let xnβ be a root in the interval (n,n+1). In this interval, βxnββ=n and {xnβ}=xnββn, so we must have axn2β=nxnββn2, i.e., axn2ββnxnβ+n2=0. We can homogenize this equation by setting xnβ=nΞΆ; then x1β=ΞΆ, and ΞΆ is a root of aΞΆ2βΞΆ+1=0.
Suppose N is the largest integer for which there is such a root; we have, for n=1,2,β¦,N,
n<xnβ=nΞΆ<n+1
Summing over nβ{1,2,β¦,N} we get
21βN(N+1)<420=21βN(N+1)ΞΆ<21βN(N+3)
From the right inequality we get 27<N and from the left one we get N<29. Thus N=28. Using this in the middle equality we get ΞΆ=2930β. Since ΞΆ satisfies aΞΆ2βΞΆ+1=0, we get
a=ΞΆβ2(ΞΆβ1)=302292ββ 291β=90029β .
The answer is 29+900=(C)929β .
OR
First note that βxββ {x}<0 when x<0 while ax2β₯0βxβR. Thus we only need to look at positive solutions ( x=0 doesn't affect the sum of the solutions). Next, we break βxββ {x} down for each interval [n,n+1 ), where n is a positive integer. Assume βxβ=n, then {x}=xβn. This means that when xβ[n,n+1), βxββ {x}=n(xβn)=nxβn2. Setting this equal to ax2 gives
nxβn2=ax2βΉax2βnx+n2=0βΉx= \dfrac{n \pm \sqrt{n^{2}-4 a n^{2}}}
We're looking at the solution with the positive x, which is x=2anβn1β4aββ=2anβ(1β1β4aβ). Note that if βxβ=n is the greatest n such that βxββ {x}=ax2 has a solution, the sum of all these solutions is slightly over βk=1nβk=2n(n+1)β, which is 406 when n=28, just under 420 . Checking this gives