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.
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.
To check if a number is divisible by another number, we divide. When we divide, there cannot be a remainder. For example, is 9 divisible by 3? Yes, it is because 9 divided by 3 is 3 without a remainder. How about 4? No, because when we divide 4 by 3 there is a remainder of 1. nycfunction@yahoo.com sum of the numbers in the given digits should be divisible by 3.then we can say that it is divisible by 3.example :22344 is divisible by 3.since2+2+3+4+4=15.so the given digit is divisible by 3
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
you take your remainder and divide it by your divisor 9 divided by 4 = 2, remainder 1 the divisor is 4, so 1/4 = .25 therefore 2 + .25 = 2.25
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.
26. This is because when you divide 26 by 5, you have a remainder of 1, and when you divide it by 4, you have a remainder of 2
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.
23.75