0
What else can I help you with?
When you divide by 4 are the only remainders you can get are 1 2 or 3?
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.
What remainders can you get when you divide by 3?
2, 1 or 0.
What is the sum of the possible remainders if the divisor is 5?
10.
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
How many remainders are possible when 9 is made are the divisor?
When 9 is used as a divisor, the remainders can range from 0 to 8. This is because the remainder is always less than the divisor. So, if you divide any number by 9, the possible remainders can be 0, 1, 2, 3, 4, 5, 6, 7, or 8.
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.
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.
How many are possible nonzero remainders by 3?
In division by three, possible nonzero remainders are 1 and 2.
How do you divide the alphabet into 5 groups?
One way to divide the alphabet into five groups is by using the concept of modular arithmetic. With the English alphabet having 26 letters, we can assign each letter a numerical value (A=0, B=1, ..., Z=25). We then divide these numerical values by 5 and group the letters based on the remainder. Group 1 would consist of letters with remainders of 0, Group 2 with remainders of 1, and so on until Group 5 with remainders of 4. This method ensures an equal distribution of letters across the five groups.
What is the greatest remainder you can have when you divide by 5?
1
