#Playing with numbers-3

Computer Science Level 3

A function F b ( x ) F_b(x) gives the sum of the digits of a decimal non-negative integer x x after converting it into base b b . Such as F 2 ( 6 ) = 2 F_2(6)=2 ,the sum of the digits of the binary of 6,110. Find the lowest 5 digit number X X for which F 3 ( X ) = 14 F_3(X)=14 and F 7 ( X ) = 20 F_7(X)=20 .

Here are the previous problems of the series :

#Playing with numbers-1

#Playing with numbers-2


The answer is 10610.

Rafsan Rcc by Rafsan Rcc , 17, Bangladesh

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3 solutions

Rafsan Rcc
May 10, 2021

I used the following codes 

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#include<stdio.h>
#include<math.h>

unsigned long int convert(unsigned long int number,int base);
unsigned long int F(unsigned long int number,int base);

int main()
{   unsigned long int i,q;
    i=10000;
    q=1;
    do{
        if(F(i,3)==14 && F(i,7)==20)
        {printf("%lu\n",i);
         q=0;}
        i+=1;
    }
    while(q!=0);
}

//convert function takes a decimal number and converts it to another base
unsigned long int convert(unsigned long number,int base)
{
    unsigned long int n,result;
    int i;
    i=0;
    n=number;
    result=0;
    do{result+=(n%base)*pow(10,i);
    n-=(n%base);
    n/=base;
    i+=1;}
    while(n>0);
    return result;
}

//The function F is similar to the one given in the question
unsigned long int F(unsigned long int number,int base)
{   unsigned long int result,n,i,x;
    result=0;
    x=convert(number,base);
    n=x;
    for(i=0;i<=log(x);i++)
    {result+=n%10;
    n-=n%10;
    n/=10;}
    return result;
}


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#F function finds the sum of the digits of a number converted to a base<=10
def F(number,base):
    result=[]
    n=number
    while n>0 :
        result.append(n%base)
        n//=base
    sum=0
    for i in result:
        sum+=i
    return sum

i=10000
q=1
while q!=0:
    if(F(i,3)==14 and F(i,7)==20) :
        print(i)
        q=0
    i+=1

1 Helpful 0 Interesting 1 Brilliant 0 Confused
Kunal Gupta
May 16, 2021 
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def convert_to_base_b_and_return_sum(x, b):
    ans = 0
    while x!=0:
        ans+=x%b
        x = x//b
    return ans

x = 10000
while x<1000000:
    if convert_to_base_b_and_return_sum(x, 3) == 14 and convert_to_base_b_and_return_sum(x, 7) == 20:
        break
    x+=1
print(x)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Fletcher Mattox
May 10, 2021 
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from baseconvert import base
n = 10000
while not (sum(base(n, 10, 3)) == 14 and sum(base(n, 10, 7)) == 20):
    n += 1
print(n)


1
 
10610

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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