一个多位数,把个位数调到第一位,组成的数是原来的2倍,这个数是?

1个回答

  • 以下t>0

    (210526315789473684共t个)=(105263157894736842共t个)*2

    (315789473684210526共t个)=(157894736842105263共t个)*2

    (421052631578947368共t个)=(210526315789473684共t个)*2

    (526315789473684210共t个)=(263157894736842105共t个)*2

    (631578947368421052共t个)=(315789473684210526共t个)*2

    (736842105263157894共t个)=(368421052631578947共t个)*2

    (842105263157894736共t个)=(421052631578947368共t个)*2

    (947368421052631578共t个)=(473684210526315789共t个)*2

    (105263157894736842共t个)=(052631578947368421共t个)*2

    (这个答案要用特别理解,要在首位添上0)

    设原数有n+1位,前n位数字组成的数记为a,个位为b.

    依题意,(10a+b)*2=10^n*b+a,即(10^n-2)b=19a

    因为a,b13,13-6->7,14,14-5>9,18,18-1>17,17-2>15,15-4>11,22-19>3,6,12,24>5,10