﻿ what number can be divided by 2 3 4 5 6 with a remainder of 1

# what number can be divided by 2 3 4 5 6 with a remainder of 1

For instance, if the input to an algo-rithm is a graph, the input size can be described by the numbers of vertices and edges in the graph.Given a well-dened notion of the remainder of one integer when divided by an-other, it is convenient to provide special notation to indicate equality of remainders. Work it out: If you divide it by 2,3,4, 5, or 6 you will get a remainder of 1. If you divide it by 11, the remainder will be 0 and no driver has a company number that meets these requirements and is smaller than this one." First divide the numerator by the denominator, (24/10 2 remainder 4). Write the answer out as 2 4/10.Because the denominator (5) cannot be divided evenly by the numerator (2), and it is a prime number, we know that it cannot be reduced further. For example, for n 4, the numbers are 1, 5, 9, 13, .If these numbers are divided by 4, we see that all leave a remainder of 1. So any number of the form qn 1 leaves a remainder of 1 when divided by n. The single repeating digit is 3. For 1/7, its obvious that 9/7 has a remainder, as does 99/7 (and for k3, 4 or 5 as well).Unlike 1/3, the fraction 1/7 is a special case where k p - 1, and we call the digits which repeat a cyclic number. For 1/11, k 2, since 11 divides evenly into 99. However, when divided by 11 there is no remainder. What is my house number?By dividing 781 from 2, 3, 4, 5, or 6, it will leave a remainder 1. Note: To test divisibility by any number that can be expressed as 2n or 5n, in which n is a positive integer, just examine the last n digits.Another method is multiplication by 3. A number of the form 10x y has the same remainder when divided by 7 as 3x y. One must multiply the leftmost digit of Write polynomial division in the same format you use when dividing numbers.Check You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. To solve the problem, consider a Markov chain taking values in the set S i : i 0, 1, 2, 3, 4, where i represents the number of umbrellas in the place where IAn integer k 1 is divisible by 7 if it leaves remainder 0 when divided by 7. When we divide an integer k 1 by 7, the possible remainders are.

144 is the least number which when divided by the given numbers will leave remainder 0 in each case.4. We have discussed how we can find just by looking at a number, whether it is divisible by small numbers 2,3,4,5,8,9 and 11. Similar questions based upon starting with an even number, then dividing by 2 and multiplying by, say, 5 and explaining why the difference between the two answers is always a multiple of 9 (or why the sum is a multiple of 11) can be readily constructed. Note: The reason why this works is that if we have: abc and b is a multiple of any given number n, then a and c will necessarily produce the same remainder when divided by n. In other words, in 2 7 9, 7 is divisible by 7. So 2 and 9 must have the same reminder when divided by 7. 17.

Integer n is divisible by 3. When n is divided by 4, the remainder is 2. What is. the value of n if it is greater than 10 but less than 30?The prime factorization of a number that is divisible by both 6 and 4 will thus be: 223 Note that this is consistent with what we saw on the number line, as well. The numbers divided by 5 gives 1 remainder means it should end either 1 or 6. But in case of number ends with 6 will be divided by 2. So only number ending with 1 have to consider. Also the number should be divisble by 7 At the start of each chapter, a brief Table of Contents (specifying section number and titles only, e.g x5.1, x5.2) is also given for convenience.1.4 Example (function) Take X 1, 2, 3, . . . , 10 and let f be the rule that for each x X, f (x) rx, where rx is the remainder when x2 is divided by 11. The numbers which can be divided by three but leave a remainder of two are: 2, 5, 8, 11, 14With these numbers, I then tried to match a number in one category with a number in the other category which was the same, and I got the answer! 1 row with 6 chairs in the row.It is clear that any multiple of a whole number can be divided by that number with no remainder. 146 For free distribution. Let us discuss more about multiples. 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.As always we will start out with a brute force approach to it. Which is pretty simple, I make a while loop, and as long as the remainder is different from zero for one of the divisors we So we get a quotient of q(x) x2 4x 9 on top, with a remainder of r(x) 30. You know, from long division of regular numbers, that your remainder (if there is one) has to be smaller than whatever you divided by. Because 1,001 is divisible by seven, an interesting pattern develops for repeating sets of 1, 2, or 3 digits that form 6-digit numbers (leading zeros are allowed) in that all such numbers are divisible by seven.Finding remainder of a number when divided by 7. The Remainder Theorem: Suppose p is a polynomial of degree at least 1 and c is a real number. When p(x) is divided by x c the remainder is p(c). The proof of Theorem 3.5 is a direct consequence of Theorem 3.4. That is why to divide the whole of something, which is 100, into 100 equal parts -- that is, to find 1 of a number -- we divide by 100.Solution. We must calculate 35 8. Now, 8 goes into 32 exactly, but 8 does not go into 35 exactly: There is a remainder of 3. What youre looking for is a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, or 6, but that is evenly divisible by 7. So as you already noticed, you only need to look at multiples of 7. Lets try 7. We get off to a good start.