24 is answer.
3 divided by 2 has a remainder of 1. Which is 1 less than 2.
The remainder is always 31 or less.
1098765433 will have a remainder of 1.
37 is the number.
Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.
989. If there is a remainder of 2 when divided by 3, the number is one less than a multiple of 3. If there is a remainder of 4 when divided by 5, the number is one less than a multiple of 5. Thus the number required is one less than a multiple of the lowest common multiple of 3 and 5 (that is 15). So what is needed is an even multiple of 15 less than or equal to 1000: 1000 ÷ 15 = 662/3 Thus the highest even multiple of 15 not greater than 1000 is 66 x 15 = 990, and the required number is 989.
14. The largest possible number for a remainder is 1 less than the divisor.
If the remainder is greater than the divisor then you can divide it once more and get one more whole number and then have less remainders.
The largest possible number for a remainder is 1 less than the number of the divisor, so it is 5.
Yes- A remainder can be any number less than the dividend (the number by which the divisor is divided). An example of a 2 digit number is: 131/11=11 remainder 10.
23 / 7 = 3 R 2, 23 < 40, 23 / 4 = 5 R 3
No. A mixed number is greater than 1 since it has a non-zero whole number (which is at least 1) and a fraction (which is greater than 0); any number divided by a number greater than 1 will be less than the original number. So 15 divided by a mixed number will be less than, not greater than, 15.
Every number is divisible by 4. A number less than 4 gives a quotient less than 1 when divided by 4. A number greater than 4 gives a quotient greater than 1 when divided by 4. 4 divided by 4 = 1.
Nothing. The remainder has to be less than the divisor.
You will get a quotient that is a positive number less than 10.
The remainder is less than the divisor because if the remainder was greater than the divisor, you have the wrong quotient. In other words, you should increase your quotient until your remainder is less than your divisor!
x-12 (greater than symbol) x/2