Question

How to find the first 3 powers of 2 whose leading digits are 12 in C

Answer 1

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

// return true if 2^n starts with prefix, else false
int startsWithPrefix(long long n, int prefix) {
    static const double log2v = 0.30102999566398119521; // log10(2)

    double x = n * log2v;
    double frac = x - floor(x);

    // count digits in prefix
    char buf[32];
    sprintf(buf, "%d", prefix);
    int digits = strlen(buf);

    // compute leading digits
    int leading = (int)floor(pow(10.0, frac + digits - 1));

    return leading == prefix;
}

int main(void) {
    int prefix = 12;
    long long results[3] = { 0 };
    int found = 0;

    for (long long n = 1; found < 3; n++) {
        if (startsWithPrefix(n, prefix)) {
            results[found++] = n;
        }
    }

    printf("First 3 powers of 2 starting with %d:\n", prefix);
    for (int i = 0; i < 3; i++) {
        long long n = results[i];
        printf("n = %lld : 2 ^ %lld = %.10g\n", n, n, pow(2.0, (double)n));
    }

    return 0;
}


/*

First 3 powers of 2 starting with 12:
n = 7 : 2 ^ 7 = 128
n = 80 : 2 ^ 80 = 1.20892582e+24
n = 90 : 2 ^ 90 = 1.237940039e+27

*/

 

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