Find the minimum of the value such $15625|(1024x-8404)$

Question

Find the minimum of the value $x\in N^{+}$ such that $$15625\mid(1024x-8404)$$

I have found that when $x\le 50$, the condition is not satisfied.

How can I find this $x$ by hand?

Answer 1

You certainly have heard about the Euclid division algorithm. What you are asked to find is $x, y$ such that $ 15625 y = 1024 x - 8404 $, or equivalently, $$15625 y + 1024 x = 8404 $$, which exists iff $( 15625, 1024 ) \mid 8404 $, which in fact holds because $( 15625, 1024 ) = 1$. So we are on the right path.

All we have to do is to solve the previous equation in it's full generality (find all solutions) and then find the one that makes $x$ the smallest but positive.

For that, we will find a particular solution with the Euclid algorithm $(x', y')$, and will solve the homogeneous equation to obtain $(x_2, y_2 )$ (this equation will be easy to solve). This will tell us that all the solutions are of the form $(x', y') + ( x_2, y_2)$ where $(x_2, y_2)$ runs all the solutions of the homogeneous equation.

So all you have to do is to find $$15625 y_1 + 1024 x_1= 1 $$ by the Euclid algorithm (which will give you $y_1 = 313$ and $x_1 = -4776 $, you should check!!) and set the particular solution $(x', y') = 8404 ( x_1, y_1)$, so $x' = 8404 x_1 = - 8404 \times 4776 $

As you want positive solutions for $x$ all you have to do as you have a particular solution, is to solve the homogeneous equation and work with that, i.e. solve $$15625 y_2 + 1024 x_2= 0 $$ and any solution of the first equation will come as $ ( x' + k x_2 , y' + k y_2 )$. As you can find, since $( 15625, 1024 ) = 1$, all solutions of $(x_2, y_2)$ are of the form $K ( 15625, -1024 ) $, so you have

$$(x, y) = ( x' + K 15625, y' - K 1024 )$$

So $x= x' + K 15625 = - 8404 \times 4776 + K 15625 $, all you have to do is to find the suitable (smallest) $K$ that makes $x$ positive, so $ x > 0 \Rightarrow x'+ K 15625 > 0 \Rightarrow K > \frac{- x' }{ 15625 }$, so put $K = \lceil \frac{ -x' }{15625}\rceil$ and $x= x'+ K 15625$ will give you the right solution.

Answer 2

$$\color{red}{\bf 3121}$$ Let $a=15625$, $b=1024$ and $c=8404$, then $a=5^6$ and $b=2^{10}$ hence $\gcd(a,b)=1$. Euclid's algorithm yields a mechanized way of finding a pair of integers $(n,m)$ such that $$an-bm=1.$$ Then, $$a(nc)-b(mc)=c.$$ Furthermore, using again that $\gcd(a,b)=1$, one knows that every solution $(i,j)$ of the identity $ai-bj=0$ can be written as $(i,j)=(bk,ak)$ for some integer $k$. Thus, the solutions $(i,j)$ of the identity $$bj-ai=c$$ are $$(i,j)=(bk-nc,ak-mc),\qquad k\in\mathbb Z.$$

Since $a\mid bx-c$ if and only if $bx-ay=c$ for some integer $y$, the answer is the smallest positive integer $$j=ak-mc.$$ It happens that $n=313$ and $m=4776$ hence $$ak-mc=k\cdot15625-4776\cdot8404$$ is positive minimal for $$k=\left\lceil\frac{4776\cdot8404}{15625}\right\rceil=2569,$$ then $$j=ak-mc=2569\cdot15625-4776\cdot8404=3121.$$

Answer 3

We need $15625|(1024x-8404)$ and $1024x-8404=4(256x-2101)$

As $(15625,2)=1;15625$ must divide $256x-2101=256(x-8)-53$

$\implies x-8\equiv(256)^{-1}\cdot53\pmod{15625}\ \ \ \ (1)$

Like Solving a Linear Congruence,

$$\frac{15625}{256}=61+\frac9{256}=61+\frac1{\dfrac{256}9}$$

$$=61+\frac1{28+\dfrac49}=61+\frac1{28+\dfrac1{\dfrac94}}=61+\frac1{28+\dfrac1{2+\dfrac14}}$$

The previous convergent of $\dfrac{15625}{256}$ is $61+\dfrac1{28+\dfrac12}=\dfrac{3479}{57}$

So, $57\cdot15625-256\cdot3479=1$ $\implies-256\cdot3479\equiv1\pmod{15625}\implies(256)^{-1}\equiv-3479$

Using $(1), x-8\equiv(-3479)\cdot53\pmod{15625}\equiv3113$