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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.

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Q: When you divide by 4 are the only remainders you can get are 1 2 or 3?
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Related questions

When you divide by 3 are the only remainders you can get are 1 or 2?

1 & 2 are the only non-zero remainders you can get from dividing a whole number by 3.


What remainders can you get when you divide by 3?

2, 1 or 0.


How can you get remainders of one or minus one when you divide an uneven integer by two?

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)


When you divide a 2 digit number by 9 and 8 you get 3 remainders?

It is 75.


What is the process of converting a decimal number into its binary equivalent?

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


What causes a repeating decimal in the long division algorithm?

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.


What causes a repeating decimal in the long division?

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.


How many are possible nonzero remainders by 3?

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.


Is there is possible alogrithm for converting a decimal to binary representation in terms of speed?

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.


How many remainders are possible when 8 is made as divisor?

The possible remainders are {0, 1, 2, 3, 4, 5, 6, 7} making eight of them.


What are the even numbers from 1 to 200?

They are the numbers that when divided by 2 have no remainders


How many possible remainders are there if 4 is the divisor?

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)