cubic power for each digit

Wed May 27, 2009 12:34 am

i an trying to solve a problem that has a no. "abc", if we make the cubic power for each digit of the number, then add the results, it should give the same number "abc" as above
im using the C language and need some help to execute my program
here are the codes i used :

Code:: include<stdio.h> int main () { int pow (int x, int y); int i,j,k; int sum=0; int c,d,f; for(i=1;i<5;i++) {for(j=1;j<5;j++) {for(k=1;k<5;k++) { c==i^3; d==j^3; f==k^3; }}} sum=sum+c+d+f; if(sum=i*100+j*10+k) {printf("the %d %d %d are numbers having the sum of cubic power of each is %d\n\n\n",i,j,k,sum); } return 0; }

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Mon Aug 24, 2009 4:50 pm

I really don't understand your problem .But it is seems interesting can you clear it more .

Sun Aug 30, 2009 8:25 pm

what compiler you are using in order to understand and help you to execute your programme each one has its own execution methods in some you might need to add some specified libraries;)

Thu Nov 04, 2010 4:04 pm

Hi cutedevils,

quite late, but here it is.

To face the picture: You want to know, which three-digit-number is equal to the sum of the cubic-power of its digits. so abc = a^3 + b^3 + c^3.

A few words to C/C++.

Neither C nor C++ are having a power operator ^ like Basic. But the operator ^ does exist. It is the XOR operator.
The line c == i^3 is syntactically correct but does nothing. Its meaning is: compare c to (i XOR 3). But the result of this comparison is discarded because it isn't assigned to any variable.
Your summing code is at the wrong position. It is executet after all loops. It will only be executed for the i = 4, j = 4 and k = 4.
The comparison if(sum=i*100+j*10+k) will always evaluate to true. Because the comparison is done with "==". The meaning of your line is: assign (i*100+j*10+k) to s and test if s is unequal to zero. Yes, the if statement is quite confusing in C/C++.

A corrected version looks like this:

Code:: #include <stdio.h> int main() { int i, j, k; int c, d, e; int sum = 0; for (i = 1; i < 10; i++) { for (j = 1; j < 10; j++) { for (k = 1; k < 10; k++) { c = i*i*i; d = j*j*j; e = k*k*k; sum = c + d + e; if (sum == (i*100 + j*10 + k)) { printf("%d%d%d = %d^3 + %d^3 + %d^3 = %d + %d + %d\n", i,j,k,i,j,k,c,d,e); } } } } return 0; }

The following output will be done:

Code:: 153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27 371 = 3^3 + 7^3 + 1^3 = 27 + 343 + 1

Best regards

Wed Mar 21, 2012 5:25 pm

Did you fix the issues?

cubic power for each digit

cubic power for each digit

Re: cubic power for each digit

Re: cubic power for each digit

Re: cubic power for each digit

Re: cubic power for each digit

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