999 + 999/999
999 + (9/9) = 1000
About 111 times
111.1111 x 9
5 times
The sum 9 + 99 + 999 + 9999 + 99 999 + .... where the last number to be added consists of nine digits of 9 becomes: (10 – 1) + (100 – 1) + (1000 – 1) + (10 000 – 1) + ... where the last multiple of 10 has nine zeros. There are nine pairs of brackets altogether. Add the multiples of 10 first and then subtract 9: The digit 1 appears 9 times in the final answer.
999 + (9/9) = 1000
One way could be: 999 +9/9 -9/9 +9/9 = 1000
9 x 1000 = 9000
To determine how many times 9 goes into 1000 evenly, you can perform the division (1000 \div 9). This results in approximately 111.11, which means 9 goes into 1000 a total of 111 times with a remainder. Therefore, 9 goes into 1000 evenly 111 times, with some left over.
It is 1000 times greater because 1000 times 0.09 = 90
About 111 times
999 + 999/999 = 1000
111.1111 x 9
5 times
If you mean as in 90.09 then it is 1000 times greater because 1000*0.09 = 90
111 times.... with 1 remaining.
A five digit number with 9 is 99999. A number which is 1000 times larger (not 1000 times as large) is 99999000.