The smallest positive integer, p, that satisifes
p = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.
So the solution set is 58 + 360k where k is an integer.
The smallest positive integer, p, that satisifes
p = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.
So the solution set is 58 + 360k where k is an integer.
The smallest positive integer, p, that satisifes
p = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.
So the solution set is 58 + 360k where k is an integer.
The smallest positive integer, p, that satisifes
p = 5a + 3 = 8b + 2 = 9c + 4 (where a, b and c are integers) is 58.
So the solution set is 58 + 360k where k is an integer.
5 */. 3= 1 Remainder:2
If you take any four consecutive numbers and divide them by 3, the remainders are as follows: 9/3 = 3 10/3 = 3 remainder 1 11/3 = 3 remainder 2 12/3 = 4 Therefore, the highest remainder you can have by dividing a whole number is 2.
Quotient: 2x+3 Remaider: 2
87 is one of infinitely many numbers that meet these requirements.
5.5
When you divide by 5, the remainder can be 0, 1, 2, 3, or 4, but no more than 4.
5 */. 3= 1 Remainder:2
4.6667
4.6667
58
Yes - the remainder can be either 1, 2, 3, 4, 5, 6, 7, or 8.
If you take any four consecutive numbers and divide them by 3, the remainders are as follows: 9/3 = 3 10/3 = 3 remainder 1 11/3 = 3 remainder 2 12/3 = 4 Therefore, the highest remainder you can have by dividing a whole number is 2.
20 / 4 = 5 (no remainder) 21 / 4 = 5 remainder 1 22 / 4 = 5 remainder 2 23 / 4 = 5 remainder 3 24 / 4 = 6 (no remainder) Therefore the numbers that can remain after you've divided a number by four are 0, 1, 2, and 3.
23.75
You divide by the denominator. You write the result of the division as the whole number, and the remainder on top of the old denominator. Example: 14/3; divide 14 by 3. The result is 4, remainder is 2, so you write the result as 4 2/3.
Well, darling, when you divide any number by 5, the largest remainder you can get is 4. Why? Because when you divide by 5, the remainders can only be 0, 1, 2, 3, or 4. So, if you wanna keep it simple and sassy, the largest remainder with a divisor of 5 is 4.
Quotient: 2x+3 Remaider: 2