What is $b $?

haha sorry. i meant 5

Ignore it

I am getting total 6 solutions

@mathzoo integers are not asked real nos are asked @the_excecutioner please list them

How? 8char.

got 2 solutions ............ $ 3+\sqrt3 $ and $ 3+\sqrt2$ though more solutions are possible...but i cannot find a fault in my work...will shortly post it

i got 5 ---- $3+\sqrt(3/2) , 3+\sqrt2 , 3+\sqrt(5/2) , 3+\sqrt3 , 3+\sqrt(7/2)$

How to solve this question is a = [a] + {a} and [a] = 4.. now get the resulting quadratic and differentiate to get minima, maxima in (0,1) . You get a few irrational solutions

kgdpsdurg wrote: @mathzoo integers are not asked real nos are asked @the_excecutioner please list them dear kgdpsdurg, plz quote the persons u are concerned with. thanku

@TheOneYouWant sorry but approach that does not involve calculus is encouraged

Use completing the squares. It's the same thing, really

{a}=a-4 This gives a quadratic if we let the integer in question to be k. Now get some bounds on k using bounds on a. This gives the possible values of k and hence lets us solve for a.

WizardMath wrote: {a}=a-4 This gives a quadratic if we let the integer in question to be k. Now get some bounds on k using bounds on a. This gives the possible values of k and hence lets us solve for a. how many real a do u get??/ ...if u gave rmo...how many did u solve

i solved 5 i misread prob 5

I gave gmo, but the bounds on a were 3<a<4 So i got 4 solutions

if the integer is k, we get the bounds as -42<=k<=24. Sorry for the inconvenience. I cannot Latex it.

Here is my solution For all k in the range we get real a

Attachments:

My general solution is $3+\frac {1}{\sqrt2} \sqrt{18-p} $, where $p $ is an integer such that $10 <p <16$.

kgdpsdurg wrote: i got 5 ---- $3+\sqrt(3/2) , 3+\sqrt2 , 3+\sqrt(5/2) , 3+\sqrt3 , 3+\sqrt(7/2)$ i am getting the same answer.

@Mathzoo another way to say the same thing

pjmaths wrote: kgdpsdurg wrote: @mathzoo integers are not asked real nos are asked @the_excecutioner please list them dear kgdpsdurg, plz quote the persons u are concerned with. thanku sorry sir, pjmaths sir

Seems no one wants to add rmo '15 to official contest collections page

kgdpsdurg wrote: Seems no one wants to add rmo '15 to official contest collections page It will probably be created after 20 th december after Chennai RMO which is postponed due to floods

the answer to question 6: http://www.artofproblemsolving.com/community/c6h1171210

I got the generalised version in the delhi rmo were we had to prove that an infinite no of real numbers a exist which satisfy the same condition of your problem.I proved that any real number with its gif an odd integer and fractional part 0.5 would be a solution . Note in general we need to find integers x mod a sufficient power of 10 such that x=3x mod 10^k 5 certainly is a solution for k=1 thus 0.5 we cant have no ending with 0 so all such decimals or fractional parts must end with 5 thus +-0.5 is a solution i investigated other cases but found no other solutions does a proof exist for this or fora sufficiently large x and k a solution exists of course other solutions suchas root of 3/2 and 7/2 do exist . I was only specifying for rational fractional parts my modular thoery

Note that in my problem a(a-3(a)) was an integer

WizardMath wrote: Here is my solution For all k in the range we get real a I think there is something wrong in Your answer. When you open up the brackets.

Here's my solution: $\{a\}=a-4$ $a(a-3\{a\})=-2a^2+12a=k \text{ (say)}$ $\implies 2a^2-12a+k=0$ $\implies a = 3 \pm \sqrt{\frac{18-k}{2}} $ $ 4 < a < 5 $ $ \implies 1 < \sqrt{\frac{18-k}{2}} < 2 $ $ \implies 16 > k > 10 $ $ \implies k \in \{11,12,13,14,15\} $ Substituting the 5 values of $k$ in the quadratic formula gives 10 solutions, but 5 of them are to be ignored because they don't lie between $4$ and $5$. $ a \in \left\{ 3 + \sqrt{\frac{7}2}\right\} \cup \left\{3 + \sqrt{3}\right\} \cup \left\{3 + \sqrt{\frac{5}2}\right\} \cup \left\{3 + \sqrt{2}\right\} \cup \left\{3 + \sqrt{\frac{3}2} \right\} $