2013
11-09

# Multiply

6*9 = 42″ is not true for base 10, but is true for base 13. That is, 6(13) * 9(13) = 42(13) because 42(13) = 4 * 131 + 2 * 130 = 54(10).

You are to write a program which inputs three integers p, q, and r and determines the base B (2<=B<=16) for which p * q = r. If there are many candidates for B, output the smallest one. For example, let p = 11, q = 11, and r = 121. Then we have 11(3) * 11(3) = 121(3) because 11(3) = 1 * 31 + 1 * 30 = 4(10) and 121(3) = 1 * 32 + 2 * 31 + 1 * 30 = 16(10). For another base such as 10, we also have 11(10) * 11(10) = 121(10). In this case, your program should output 3 which is the smallest base. If there is no candidate for B, output 0.

The input consists of T test cases. The number of test cases (T ) is given in the first line of the input file. Each test case consists of three integers p, q, and r in a line. All digits of p, q, and r are numeric digits and 1<=p,q, r<=1,000,000.

Print exactly one line for each test case. The line should contain one integer which is the smallest base for which p * q = r. If there is no such base, your program should output 0.

3
6 9 42
11 11 121
2 2 2


13
3
0



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import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner in=new Scanner(System.in);
int a=in.nextInt();
while((a--)!=0)
{
String s1=in.next();
String s2=in.next();
String s3=in.next();
int max=0;
for(int i=0;i< s1.length();i++)
{
char c=s1.charAt(i);
int u=c-48;
if(max< u)max=u;
}
for(int i=0;i< s2.length();i++)
{
char c=s2.charAt(i);
int u=c-48;
if(max< u)max=u;
}
for(int i=0;i< s3.length();i++)
{
char c=s3.charAt(i);
int u=c-48;
if(max< u)max=u;
}
boolean b=true;
for(int o=max+1;o< 17;o++)
{
int w1=Integer.parseInt(s1, o);
int w2=Integer.parseInt(s2, o);
int w3=Integer.parseInt(s3, o);
if(w1*w2==w3)
{
System.out.println(o);
b=false;
break;
}
}
if(b)System.out.println(0);
}
}
}

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