The first month 1000

The second month is1000+1000 (1+1%) =1000 (1+1).

The third month is1000+1000 (1+1.01) =10000.

The fourth month is1000+1000 (1+1.012) (1+1%).

Therefore, there is1000 in month X (1+1+1.2+1.01.3+65438+).

1+ 1.0 1.0 1.2+ 1.3+ 1.0 1.4+...+65438.

1+ 1.0 1+ 1.0 1^2+ 1.0 1^3+ 1.0 1^4+……+ 1.0 1^(x- 1)

[ 1( 1- 1.0 1^x)]/( 1- 1.0 1)= 100( 1.0 1^x- 1]

I asked if I could buy an 88888 car in a few months, just to solve the inequality.

1000*[ 100( 1.0 1^x- 1)]<； =88888

1.0 1^x<； = 1.88888

So (log1.01)1.88888.

Only by borrowing high school math tools can we work out specific results.