Mad factorization

Algebra Level 4

Find a factor of 100140001 which lies between 12000 and 13000.

Note : try to solve the problem in an algebraic way.


The answer is 12181.

Choi Chakfung by choi chakfung , 28, Hong Kong

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2 solutions

Adnan Roshid
May 27, 2016

Here is a C++ code for it . 

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#include <bits/stdc++.h>

using namespace std;

int main()
{
   for(int i = 12000; i <= 13000; i++) {
      if(100140001 % i == 0) cout << i << endl;
   }
   return 0;
}

Here is the output
output: 12181 12181

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for sharing your insight, Adnan. I upvoted your solution. (+1)

Pranshu Gaba - 5 years ago
Jash Shah
Apr 18, 2016

100140001 can be written as 10^8+14*10^4+1 Let 10 be equal to x

x 8 + 14 x 4 + 1 = x 8 + 2 x 4 + 1 + 12 x 4 x^8+14x^4+1=x^8+2x^4+1+12x^4

= ( x 4 + 1 ) 2 + 4 x 2 ( x 4 + 1 ) + 4 x 4 + 8 x 4 − 4 x 2 ( x 4 + 1 ) ={(x^4+1)^2+4x^2(x^4+1)+4x^4}+8x^4-4x^2(x^4+1)

= ( x 4 + 2 x 2 + 1 ) 2 − ( 2 x 3 − 2 x ) 2 =(x^4+2x^2+1)^2-(2x^3-2x)^2

Using identity we get ( x 4 + 2 x 3 + 2 x 2 − 2 x + 1 ) ( x 4 − 2 x 3 + 2 x 2 + 2 x + 1 ) (x^4+2x^3+2x^2-2x+1)(x^4-2x^3+2x^2+2x+1)

Substituting x = 10 x =10 we get 100140001 = 12181 × 8221 100140001=12181 \times 8221

1 Helpful 0 Interesting 1 Brilliant 0 Confused

exactly what i did

choi chakfung - 5 years, 1 month ago

Same method

Aditya Kumar - 5 years ago

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