Twenty times.
To find how many times the digit '9' appears between 1 and 1100, we can analyze each digit place (hundreds, tens, and units) separately. In the range from 1 to 999, the digit '9' appears 300 times: 100 times in the units place (from 9, 19, ..., 999), 100 times in the tens place (from 90-99, 190-199, ..., 990-999), and 100 times in the hundreds place (from 900-999). In the range from 1000 to 1100, '9' does not appear. Therefore, the total count of '9's from 1 to 1100 is 300.
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
With repeats: 4×4 = 16; Without repeats: 4×3 = 12.
Well if repeats are allowed, then unlimited. If you can repeat the numbers then the list would go on forever.
how many times gerater is the underlined digit 5 in 365,486,201 than the digit 5
120
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
Since 1100 divided by 9 is 122.222... , 9 can go into 1100 122 times.
With repeats: 4×4 = 16; Without repeats: 4×3 = 12.
1.1936 times.
Well if repeats are allowed, then unlimited. If you can repeat the numbers then the list would go on forever.
55 times.
Exactly 44 times
if the repetition is allowed the there is 6*6*6 possible ways = 216
how many times gerater is the underlined digit 5 in 365,486,201 than the digit 5
36.67 times, or 36 times with a remainder of 20.
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.