9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92
Eleven.
9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99 =20
Well there will be 1 nine for all the 9s in the units column (9, 19, 29 etc.), and 1 nine for all the numbers with 9 in the tens column (90, 91, 92 etc.). So the answer is 20.
111 of them with a remainder of 1
There are: 640/9 = 71 with a remainder of 1
Eleven.
9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99 =20
Well there will be 1 nine for all the 9s in the units column (9, 19, 29 etc.), and 1 nine for all the numbers with 9 in the tens column (90, 91, 92 etc.). So the answer is 20.
111 of them with a remainder of 1
There are: 640/9 = 71 with a remainder of 1
8, remainder 1
1 or 4 , depending on your point of view.
11 with 1 remaining 100 - 1 = 99 = 9 x 11
1 ns (nanosecond) = 10-9s 1s / 10-9s = 109 109 = 1000 000 000 1 second = 1000 000 000 ns (nanosecond)
Only 1 exists, and it is "999"
To find the whole numbers between 62 and 93, you exclude the endpoints. The whole numbers start from 63 and go up to 92. To calculate the count, subtract 63 from 92 and add 1: (92 - 63 + 1 = 30). Therefore, there are 30 whole numbers between 62 and 93.
Including the start and end date in the calculation, there are 92 days between the two dates.