2009的所有自然数中,有多少个整数X使2x与X2被7除余数相同

2个回答

  • X是一位数时

    2X = 20+X = 21 + (X-1)

    X2 = 10X + 2 = 7X + (3X+2)

    则3X + 2 = X-1 + 7K

    2X = 7K - 3,显然K = 1、3,X = 2、9

    X是二位数时

    2X = 200+X = 196+ (X+4)

    X2 = 10X + 2 = 7X + (3X+2)

    则X+4 = 3X + 2 - 7K

    2X = 7K + 2,显然K是偶数K=2T,X = 7T+1,即X是被7除余1的二位数:

    X从15到99共13个.

    X是三位数时

    2X = 2000+X = 2002+ (X-2)

    X2 = 10X + 2 = 7X + (3X+2)

    则X - 2 = 3X + 2 - 7K

    2X = 7K - 4,显然K是偶数K=2T,X = 7T-2,即X是被7除余5的三位数:

    X从103到999共129个.

    X是四位数时

    2X = 20000+X = 19999+ (X+1)

    X2 = 10X + 2 = 7X + (3X+2)

    则X + 1 = 3X + 2 - 7K

    2X = 7K - 1,显然K是奇数K=2T+1,X = 7T+3,即X是被7除余3的小于等于2009的四位数:

    X从1004到2005共144个.

    综上,共有2+13+129+144=288个.