the max remainder you can have when dividing by a number is that number minus 1 So 4 can only have 1, 2 and 3 as remainders. 9 can only have 1-8 and so on.
The possible remainders are {0, 1, 2, 3, 4, 5, 6, 7} making eight of them.
Only 3 non-zero remainders.1, 2, and 3 are the only possible non-zero remainders since any number greater than or equal to the divisor could also be divided, to result in a new quotient. A remainder of zero, means that the dividend is divisible by the divisor (the divisor is a factor of the number)
In division by 5, you can have remainders of 0, 1, 2, 3, and 4. If you count zero, then you can have five possible remainders. If you are not counting zero, then 4 possible remainders.
There are 9 possible remainders: 0, 1, 2, 3, 4, 5, 6, 7, 8.
With the divisor (the number you are dividing by) as 9, there are 9 possible remainders: 0, 1, 2, 3, 4, 5, 6, 7, and 8.
1 & 2 are the only non-zero remainders you can get from dividing a whole number by 3.
2, 1 or 0.
Any uneven integer can be represented in the form 2n+1. When you divide this number by 2, you will get n with 1 remaining (either +1 or -1)
It is 75.
Divide by the number repeatedly by two (until it is zero) and collect the remainders. For example: 13 / 2 = 6 Remainder 1 6 / 2 = 3 Remainder 0 3 / 2 = 1 Remainder 1 1 / 2 = 0 Remainder 1 Reading remainders from bottom yields: 1101
When you divide by a divisor q, the remainders can only be integers that are smaller than q. If the remainder is 0 then the decimal is terminating. Otherwise, it can only take the values 1, 2, 3, ... ,(q-1). So, after at most q-1 different remainders you must have a remainder which has appeared before. That is where the long division algorithm loops back into an earlier pattern = repeating sequence.
When you divide by a divisor q, the remainders can only be integers that are smaller than q. If the remainder is 0 then the decimal is terminating. Otherwise, it can only take the values 1, 2, 3, ... ,(q-1). So, after at most q-1 different remainders you must have a remainder which has appeared before. That is where the long division algorithm loops back into an earlier pattern = repeating sequence.
There are two possible nonzero remainders when dividing a number by 3: 1 and 2. Any nonzero integer can be divided by 3 resulting in either a remainder of 1 or 2.
Repeatedly divide by 2. The remainders - in reverse order - form the binary number. You must continue dividing until the result of the division is zero. Example: Convert 11(decimal) to binary. 11 / 2 = 5 r 1 5 / 2 = 2 r 1 2 / 2 = 1 r 0 1 / 2 = 0 r 1. Now list the remainders from bottom to top: 1011. This is the binary representation of eleven.
The possible remainders are {0, 1, 2, 3, 4, 5, 6, 7} making eight of them.
They are the numbers that when divided by 2 have no remainders
Only 3 non-zero remainders.1, 2, and 3 are the only possible non-zero remainders since any number greater than or equal to the divisor could also be divided, to result in a new quotient. A remainder of zero, means that the dividend is divisible by the divisor (the divisor is a factor of the number)