ok..for this question you will need to tell me the process too:
let x( x is a multidigit number) be a palindrome number. Find the least x such that x(x+1) is also a palindrome number. Any valid process will be given a medal...merely posting answers won't do

Question

ok..for this question you will need to tell me the process too:
let x( x is a multidigit number) be a palindrome number. Find the least x such that x(x+1) is also a palindrome number. Any valid process will be given a medal...merely posting answers won't do

amriju · Answer

Process need not be elaborate..just the hint will do..or the algo

anonymous · Answer

since we have any 1digit number is a palindrome number, so that you can have x =1 
and it satisfies 1(1+1) =2, right? That is the "least" x , right?

amriju · Answer

palindrome my friend...x has to be a multi digit number...you know what a palindrome number is?

anonymous · Answer

Actually, I don't know what it is but your problem is interesting so that I made a short research about it. And I know what it is , how it works http://en.wikipedia.org/wiki/Palindromic_number However, if x is multidigit number, let me work more. Hopefully, I can figure out something or at least I learn something new.

amriju · Answer

its quite a good question you know...damn interesting one...if you are interested in number theory

anonymous · Answer

if it is number theory, then @ganeshie8 is the best

anonymous · Answer

By testing, I got 77*(77+1) =6006 , that is the solution, but not know how to put it in logic yet. hihihi..However, we can derive the process from the result, right? (hopefully)

amriju · Answer

@ganeshie8 ...can u solve this?

amriju · Answer

@OOOPS ...the answer is easy..the process is what matters

anonymous · Answer

@amriju  yes, you are right, let wait for him; he is asleep; just few more hours, he wakes up and help us. hihihi

puzzler7 · Answer

A brute force method is a legit process to do this.

amriju · Answer

nope...thats the beauty..u dont need brute force..:)

puzzler7 · Answer

But it is a legit method, just not the best.

amriju · Answer

solve it mathematically..not by those methods...i want that..brute force will be merely testing each value..

anonymous · Answer

If we restrict $x$ to be a two-digit number, you can write
$$x=10x_2+x_1$$
where $x_1$ and $x_2$ are integers between 0 and 9 (including 0 and 9, except $x_2\not=0$).

For $x$ to be palindromic, you must have $x_1=x_2$, so $x=11x_2$.

From here you can brute force it and determine that $x_2=7$, or $x=77$ is the first palindrome to yield a palindromic $x(x+1)$, but we want a more elegant method, yes? I can't promise perfection, but I'll think about this problem...

amriju · Answer

even from there you dont need a brute force..

amriju · Answer

I'll post the answer if you want me to..one of the process that requires no prerequisites...

anonymous · Answer

I tried going about this by letting $y=x(x+1)$, then either
$$y=100y_3+10y_2+y_1~~~~\text{or}~~~~y=1000y_4+100y_3+10y_2+y_1$$
Equating this with the $x$ expression,
$$(10x_2+x_1)(10x_2+x_1+1)=100y_3+10y_2+y_1\~~~~~~~~~~~~~~~~~~~~~~~~\text{or}\
(10x_2+x_1)(10x_2+x_1+1)=1000y_4+100y_3+10y_2+y_1$$
depending on the value of $x_1$. Then I tried solving for $x_1=x_2$ by matching up the digits places. No luck though, unless there's something I'm not seeing.

ganeshie8 · Answer

for the last case,   $\large x_1 = x_2, y_4 = y_1, y_3=y_2$ give us a diaphontine equation :
$$\large   x_1(11x_1+1) = 91y_1+10y_2$$

solving this also gives us cases

ganeshie8 · Answer

so the easy way is to bruteforce and test the 8 cases : 11,22,..,88