How to get an "Impossible" problem right???

• Step 1 - Open Google

• Step 2 - Search "persistence of a number"

• Step 3 - Open the Wikipedia page.

• Step 4 - Find the sequence, we get $679$ as the $5^\text{th}$ number of the sequence.

• Step 5 - Type the answer.

• Step 6 - See "Correct! The answer is 679."

• Step 7 - Be happy.

$\color{#D61F06}{Was \ it \ too \ easy \ for \ an \ IMPOSSIBLE \ PROBLEM???}$

Vote your answer - Confused if Yes, and Interesting if No

I checked for all the numbers from 1.

Since there were many patterns, it took me about half an hour to find the answer :)

And my solution to this problem is, well, persistence!

$\boxed{The \ answer \ is \ \underline{679}}$

$\text{The answer is }679$

$\text{The method of getting the answer is by knowing the fact that after the first step of multiplication,}$

$\text{the number should come in the format of }2^i3^j7^k \text{ or }3^i5^j7^k$

$\boxed{i, j, k \textit{ are variables which can be equal to any number, but they should be there, otherwise it can't have persistence}}$

$\textbf{An example}$

• $679$

• $\text{After first step, it becomes }6×7×9 = 378$

• $378 = 2^13^37^1$

• $\text{In here }i = 1; j = 3; k = 1$

$\text{There isn't a general format for this question,}$

$\text{and there are conjectures on whether there is a number c where persistence never goes above }c$

$\text{But like how I said, the format I mentioned above helps find numbers with this quality using computers,}$

$\text{and not meant to be on paper}$

$\textnormal{I would assume that there is no generalization since there is no known number with persistence 12. To be honest, the way I solved}\\\textnormal{it was just putting in``10, 25, 39, 77" into OEIS. The first result was a sequence about persistence, which gave the answer of } \boxed{679}$ .

I will post a math approach, when I can.

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#include <iostream>
using namespace std;
int sol(int a, int b, int c);
int main()
{
    for(int a=0;;a++){
        for(int b=0;b<10;b++){
            for(int c=0;c<10;c++){
                if(sol(a,b,c)==5){
                    cout << a << b << c;
                    return 0;
                }
            }
        }
    }
}
int sol(int a, int b, int c){
    if (a==0&&b==0) return 0;
    if (a==0){
        int d=b*c%10, e=b*c/10;
        return sol(a,e,d)+1;
    }
    int r=a*b*c%10, t=(a*b*c-r)/10,g=t%10, l=(a*b*c-r-10*g)/100;
    return sol(l,g,r)+1;
}

$\text{\Large The number is 679}$

$\boxed{Take}$ $\boxed{it}$ $\boxed{or}$ $\boxed{leave}$ $\boxed{it}$

Here you will only find a code solution or somebody explaining how to do trial and error.

• I also did trial and error but verified from Wikipedia also.