0.7273
find common denominators, 8/11 - 4/15 = 120/165 - 44/165 = 76/165. That is your answer is simplest form.
11.0
9,867 (11 x 897)
99990 Note that if you add 11 to my answer you get a 6-digit number therefore it must be the largest 5-digit whole multiple of 11.
11
0.7273
120
-340
10.9091
0.3636
The largest 5 digit number is 99999. If this is divided by 11 it leaves a remainder of 9. Therefore the largest 5 digit number divisible by 11 is 99999 - 9 = 99990.
2divide 11
To determine the number of times the digit 1 appears between 1 and 120, we can consider each place value separately. In the units place, the digit 1 appears 12 times (1, 11, 21, ..., 111). In the tens place, the digit 1 appears 11 times (10, 11, 12, ..., 19). Therefore, the total number of times the digit 1 appears between 1 and 120 is 12 (from the units place) + 11 (from the tens place) = 23 times.
Oh, dude, yeah, totally! A remainder can definitely be a 2-digit number. It's just whatever is left over after you divide one number by another. So, like, if you divide 100 by 3, you get a remainder of 1, which is a 1-digit number. But if you divide 100 by 7, you get a remainder of 2 digits, which is totally cool too.
If the difference between the summs of every other digit can be divided by 11 or is 0. for example: 4,191: 4-1+9-1=11, 11 can be divided by 11, so 4,191 can. 876,953: 8-7+6-9+5-3=0, so 876953 can.
584.6923