Hi,

I assume you want the sum of the (decimal) digits of m. The problem is small enough that I suppose you could do all the computations by hand, but a quick little C program does all the work for you (exact same idea in any language):

Code:

#include <stdio.h>
int powers[10];
int differences[45];
int main(void) {
int i,j,k,m,sumOfDigits;
powers[0] = 2;
for (i=1;i<10;i++) {
powers[i]=2*powers[i-1];
}
k=0;
for (i=0;i<10;i++) {
for (j=i+1;j<10;j++) {
differences[k++]=powers[j]-powers[i];
}
}
m=0;
for (i=0;i<45;i++) {
m+=differences[i];
}
printf("The number m is %d\n",m);
sumOfDigits=0;
while (m!=0) {
sumOfDigits+=m%10;
m/=10;
}
printf("The sum of the digits of m is %d\n",sumOfDigits);
return(0);
}

The output of the program is m = 14362 and the sum of the digits is 16. I don't see any clever way to find these answers except by brute force computation.