[object Object]
999
3 and 333 **************** I would like to add 1 and also 9 (999 ÷ 9 = 111) to the above list.
109
108 -999, if you consider negative numbers.
100. The highest 3 digit number divisible by 9 is 999 (111 times). The highest 2 digit number divisible by 9 is 99 (11 times). The difference is 100.
The first number in this range divisible by 9 is 9 itself...9 = 1 x 9 The last number in the range is 999 = 111 x 9 So there are 111 numbers between 1 and 1000 that are divisible by 9.
No.
The largest 3-digit number is 999. To check if it is a multiple of 3, you can sum the digits: 9 + 9 + 9 = 27, which is divisible by 3. Therefore, 999 is the largest 3-digit number that is a multiple of 3.
To find the greatest 3-digit number divisible by both 5 and 9, we need to find the highest common multiple of 5 and 9 within the range of 100 to 999. The highest common multiple of 5 and 9 is 45. To find the greatest 3-digit number divisible by 45, we divide 999 by 45, which equals 22 with a remainder. Therefore, the greatest 3-digit number divisible by both 5 and 9 is 45 x 22 = 990.
That would be 999.
Between 100 and 999 there are 448.
999 + 999/999