7 digit decimal number

[David Goldberg]

Jerome beat me to the punch.  My answer was slightly smaller,
9.0e9.  Here is the program with comments that prints out
a lot of examples (include 9.0e9):

-david


/*
 * Find 7 digit decimal number d so that d rounded to 24 bits
 * (single prec) and back to 7 digits does not recover d.
 *
 *  Use fact that 2^33 =  8589934592
 *           and 10^10 = 10000000000
 *
 *   are close, so that there are more decimal numbers
 *   with 7 digits in the range (2^33, 10^10) than binary numbers
 *   with 24 bits.  Computing how many numbers in that range:
 *
 *    7 digit decimal numbers in (2^33, 10^10) are of form
 *                  i*10^3,  2^33/10^3 < i < 10^7
 *    24 bit binary  numbers   in (2^33, 10^10) are of form
 *                  j*2^10,       2^23 < j < 10^10/2^10 
 *
 *    10^7 - 2^33/10^3 =  1410065 decimal numbers
 *    10^10/2^10 - 2^23 = 1377017 binary numbers
 *
 *   Other pairs besides (2^33, 10^10) are (2^43, 10^13),
 *     (2^53,10^16), (2^63,10^19), etc.
 *
 */

#include <math.h>

main()
{
    double dlow, dhi, i, x, m;
    double d;  /* the 7 decimal number */
    double b;  /* rounded to 24 bit binary */

    dlow = ceil(scalb(1.0, 33.0)/1000.0); /* 2^33/10^3 */
    dhi = 10000000.0;                     /* 10^7 */
    m = 536870913.0;                      /* 2^k + 1 = 2^29 + 1 */
    
    for (i = dlow; i < dhi; i++) {
	d = i*1000.0;
	/* round to 24 bits, k=29 p-k=24 */
	b = x = (m*d) - (m*d - d);
	
	/* round back to 7 decimal digits */
	x = anint(x/1000.0)*1000.0;
	if (x != d) {
	    printf("%9.0f, %.0f x 2^10,  %9.0f\n",
		   d, b/1024.0, x);
	}
    }
}

/*
 *  First answer is 8589973000, 8388646 x 2^10,  8589974000
 *
 *  Checking:
 *     8589973000 is 100000000000000000100101 1000001000
 *   which rounds to 100000000000000000100110 0000000000
 *   which in decimal is 8589973 504
 *   which rounds to 8589974 000
 */